NCERT Solutions for Class 11 Maths Chapter 16 Probability

# NCERT Solutions for Class 11 Maths Chapter 16 Probability

Edited By Ramraj Saini | Updated on Sep 25, 2023 10:06 PM IST

## Probability Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 16 Probability are provided here. Provides NCERT solutions in simple, comprehensive and step by step solutions for all questions. these NCERT solutions are prepared by an expert team of Careers360 keeping in mind the latest syllabus of CBSE 2023.Students can practice these to command all the concepts which overall lead to scoring well in the board as well as in competitive exams.

There are 44 questions in 3 exercises in the NCERT textbook. First, try to solve it on your own, if you are not able to do so, you can take help of class 11 maths chapter 16 question answer. This chapter is very important for CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT etc. class 11 maths chapter 16 NCERT solutions is useful to study advanced topics like probability distribution, stochastic process, mathematics statics and probability(MSP). Here you will get NCERT solutions for class 11 also.

## Probability Class 11 Solutions - Important Formulae

Probability:

Probability is calculated as the number of favourable outcomes in an event divided by the total number of possible outcomes.

• Probability = Number of Favourable Outcomes / Total Number of Outcomes

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Event:

An event is a subset of the sample space S. An empty set is known as the Impossible event.

For any random experiment with sample space S, the probability P is a real-valued function where:

• P(E) ≥ 0 for any event E

• P(S) = 1

Mutually Exclusive Events:

If events E and F are mutually exclusive, then the probability of their union is the sum of their individual probabilities.

• P(E ∪ F) = P(E) + P(F)

Equally Likely Outcomes:

Equally likely outcomes are outcomes with the same probability of occurring.

Let S be a finite sample space with equally likely outcomes, and A be an event. The probability of event A is calculated as:

• P(A) = Number of Favourable Outcomes (n(A)) / Total Number of Outcomes (n(S))

Formulas for Combining Events:

Event P or Q corresponds to the set union P ∪ Q.

Event P and Q corresponds to the set intersection P ∩ Q.

Event P and not Q corresponds to the set difference P - Q.

Mutually Exclusive Events:

Events P and Q are mutually exclusive if their intersection is the empty set:

• P ∩ Q = φ (empty set)

Exhaustive and Mutually Exclusive Events:

Events P1, P2, ..., Pn are exhaustive and mutually exclusive if:

• Their union covers the entire sample space: P1 ∪ P2 ∪ ... ∪ Pn = S

• They have no common outcomes: Ei ∩ Ej = φ for all i ≠ j

Free download NCERT Solutions for Class 11 Maths Chapter 16 Probability for CBSE Exam.

## Probability Class 11 NCERT Solutions (Intext Questions and Exercise)

Probability class 11 ncert solutions - Exercise: 16.1

A coin is tossed three times.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

The required sample space is:

S = {HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}

Question:2 Describe the sample space for the indicated experiment.

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = { $\dpi{100} (x, y) : x, y$ = 1,2,3,4,5,6}

or S = {(1,1), (1,2), (1,3), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6, 1), (6, 2), ..., (6,6)}

Question:3 Describe the sample space for the indicated experiment.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

The required sample space is:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}

Question:4 Describe the sample space for the indicated experiment.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

And,

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Question:5 Describe the sample space for the indicated experiment.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

For H, when a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {H1, H2, H3, H4, H5, H6, T}

Question:6 Describe the sample space for the indicated experiment.

Let X denote the event Room X is selected, Y denote the event Room Y is selected.

B1, B2 denote the event a boy is selected and G1, G2 denote the event a girl is selected from room X.

B3 denotes the event a boy is selected and G3, G4, G5 denote the event a girl is selected from Room Y.

The required sample space is:

S = {XB1 , XB2 , XG1 , XG2 , YB3 , YG3 , YG4 , YG5 }

Question:7. Describe the sample space for the indicated experiment.

Let, R denote the event the red die comes out,

W denote the event the white die comes out,

B denote the event the Blue die is chosen

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

Question:8(i) An experiment consists of recording boy–girl composition of families with 2 children.

Let B denote the event a boy is born,

G denote the event a girl is born

The required sample space with a boy or girl in the order of their births is:

S = {BB, BG, GB, GG}

Question: 8(ii). An experiment consists of recording boy–girl composition of families with 2 children.

(ii) In a family with two child, there can be only three possible cases:

no girl child, 1 girl child or 2 girl child

The required sample is:

S = {0, 1, 2}

Given, Number of red balls =1

Number of white balls = 3

Let R denote the event that the red ball is drawn.

And W denotes the event that a white ball is drawn.

Since two balls are drawn at random in succession without replacement,

if the first ball is red, the second ball will be white. And if the first ball is white, second can be either of red and white

The required sample space is:

S = {RW, WR, WW}

Let H denote the event that Head occurs and T denote the event that Tail occurs.

For T in first toss, the possible outcomes when a die is thrown = {1, 2, 3 ,4 ,5 ,6}

The required sample space is :

S = {HH, HT, T1, T2, T3, T4, T5, T6}

Let D denote the event the bulb is defective and N denote the event the bulb is non-defective
The required sample space is:

S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}

Possible outcomes when a coin is tossed = {H,T}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

When T occurs, experiment is finished. S1 = {T}

When H occurs, a die is thrown.

If the outcome is odd ({1,3,5}), S2 = {H1, H3, H5}

If the outcome is even({2,4,6}), the die is thrown again.,

S3 = {H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

The required sample space is:

S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Given, two slips are drawn from the box, one after the other, without replacement.

Let 1, 2, 3, 4 denote the event that 1, 2, 3, 4 numbered slip is drawn respectively.

When two slips are drawn without replacement, the first event has 4 possible outcomes and the second event has 3 possible outcomes

S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Possible outcomes when a coin is tossed = {H,T}

If the number on the die is even {2,4,6}, the coin is tossed once.

S1 = {2H, 2T, 4H, 4T, 6H, 6T}

If the number on the die is odd {1,3,5}, the coin is tossed twice.

S2 = {1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

The required sample space is:

S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}

Possible outcomes when a coin is tossed = {H,T}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Let R1 and R2 denote the event that a red ball is drawn

and B1, B2, B3 denote the event that a blue ball is drawn

If H occurs, a die is thrown.

S1 = {H1, H2, H3, H4, H5, H6}

If T occurs, a ball from a box which contains $2$ red and $3$ black balls is drawn.

S2 = {TR1 , TR2 , TB1 , TB2 , TB3}

The required sample space is:

S = {TR1 , TR2 , TB1 , TB2 , TB3 , H1, H2, H3, H4, H5, H6}

Given, a die is thrown repeatedly untill a six comes up.

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

In the experiment 6 may come up on the first throw, or the 2nd throw, or the 3rd throw and so on till 6 is obtained.

The required sample space is:

S = {6, (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6), ..., (1,5,6), (2,1,6). (2,2,6), ..., (2,5,6), ..., (5,1,6), (5,2,6), ... }

Class 11 maths chapter 16 question answer - Exercise 16.2

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Now,

E = event that the die shows 4 = {4}

F = event that the die shows even number = {2, 4, 6}

E $\dpi{100} \cap$ F = {4} $\dpi{100} \cap$ {2, 4, 6}

= {4} $\dpi{100} \neq \phi$

Hence E and F are not mutually exclusive event.

Question:2(i) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, A : a number less than 7

As every number on a die is less than 7

A = {1, 2, 3, 4, 5, 6} = S

Question:2(ii) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, B: a number greater than 7

As no number on the die is greater than 7

B = $\dpi{100} \phi$

Question:2(iii) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, C : a multiple of 3

C = {3, 6}

Question:2(iv) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, D : a number less than 4

D = {1, 2, 3}

Question:2(v) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, E : an even number greater than 4

S1 = Subset of S containing even numbers = {2,4,6}

Therefore , E = {6}

Question:2(vi). A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, F : a number not less than 3

F = {x: x $\dpi{100} \in$ S, x $\dpi{80} \geq$ 3 } = {3, 4, 5, 6}

Question:.(vi) A die is thrown. Describe the following events:

A = {1, 2, 3, 4, 5, 6}

B= $\dpi{100} \phi$

$\dpi{100} \therefore$ A $\dpi{80} \cup$ B = { 1, 2, 3, 4, 5, 6 } $\dpi{80} \cup$$\dpi{100} \phi$ = {1, 2, 3, 4, 5, 6}

Question:2.(vi) A die is thrown. Describe the following events:

A = {1, 2, 3, 4, 5, 6}

B= $\dpi{100} \phi$

$\dpi{100} \therefore$ A $\dpi{80} \cap$ B = {1, 2, 3, 4, 5, 6} $\dpi{80} \cap$$\dpi{100} \phi$ = $\dpi{100} \phi$

Question:2.(vi) A die is thrown. Describe the following events:

B= $\dpi{100} \phi$

C= {3, 6}

$\dpi{100} \therefore$ B $\dpi{80} \cup$ C = $\dpi{100} \phi$$\dpi{80} \cup$ { 3, 6 } = {3, 6}

Question:2.(vi) A die is thrown. Describe the following events:

E = {6}

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ E $\dpi{80} \cap$ F = { 6 } $\dpi{80} \cap$ {3, 4, 5, 6 } = {6}

Question:2.(vi) A die is thrown. Describe the following events:

D = {1, 2, 3}

E = {6}

$\dpi{100} \therefore$ D $\dpi{80} \cap$ E = {1, 2, 3} $\dpi{80} \cap$ {6} = $\dpi{100} \phi$ (As nothing is common in these sets)

Question:2.(vi) A die is thrown. Describe the following events:

A = {1, 2, 3, 4, 5, 6}

C = {3, 6}

$\dpi{100} \therefore$ A - C = {1, 2, 3 , 4, 5, 6 } - { 3 , 6 } = {1, 2, 4, 5}

Question:2.(vi) A die is thrown. Describe the following events:

D = {1, 2, 3}

E = {6}

$\dpi{100} \therefore$ D - E = {1, 2, 3} - {6} = {1, 2, 3}

Question:2.(vi) A die is thrown. Describe the following events:

E = {6}

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2}

$\dpi{100} \therefore$ E $\dpi{80} \cap$ F' = {6} $\dpi{80} \cap$ {1, 2} = $\dpi{100} \phi$

Question:2.(vi) A die is thrown. Describe the following events:

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2}

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

A : the sum is greater than 8

Possible sum greater than 8 are 9, 10, 11 and 12

A = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>8 } ] = {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

B: 2 occurs on either die

Hence the number 2 can come on first die or second die or on both the die simultaneously.

B = [ {(a,b): (a,b) $\dpi{100} \in$ E, a or b = 2 } ] = {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

C: the sum is at least 7 and a multiple of 3

The sum can only be 9 or 12.

C = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>6 & a+b = 3k, k $\dpi{100} \in$ I} ] = {(3,6), (6,3), (5, 4), (4,5), (6,6)}

For two elements to be mutually exclusive, there should not be any common element amongst them.

Also, A = { (3,6) , (4,5), (5, 4), (6,3) , (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) }

B = {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}

C = { (3,6), (6,3), (5, 4), (4,5), (6,6) }

Now, A $\cap$ B = $\phi$ (no common element in A and B)

Hence, A and B are mutually exclusive

Again, B $\cap$ C = $\phi$ (no common element in B and C)

Hence, B and C are mutually exclusive

Again, C $\cap$ A = {(3,6), (6,3), (5, 4), (4,5), (6,6)}

Therefore,

A and B, B and C are mutually exclusive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(i). For two elements X and Y to be mutually exclusive, X $\cap$ Y = $\phi$

A $\cap$ B = {HHH} $\cap$ {HHT, HTH, THH} = $\phi$ ; Hence A and B are mutually exclusive.

B $\cap$ C = {HHT, HTH, THH} $\cap$ {TTT} = $\phi$ ; Hence B and C are mutually exclusive.

C $\cap$ D = {TTT} $\cap$ {HHH, HHT, HTH, HTT} = $\phi$ ; Hence C and D are mutually exclusive.

D $\cap$ A = {HHH, HHT, HTH, HTT} $\cap$ {HHH} = {HHH} ; Hence D and A are not mutually exclusive.

A $\cap$ C = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and C are mutually exclusive.

B $\cap$ D = {HHT, HTH, THH} $\cap$ {HHH, HHT, HTH, HTT} = {HHT, HTH} ; Hence B and D are not mutually exclusive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(ii).If an event X has only one sample point of a sample space, it is called a simple event.

A = {HHH} and C = {TTT}

Hence, A and C are simple events.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, THH, TTH, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(iv). If an event has more than one sample point, it is called a Compound event.

B = {HHT, HTH, THH} and D = {HHH, HHT, HTH, HTT}

Hence, B and D are compound events.

Question:5(i) Three coins are tossed. Describe

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, THH, TTH, TTT}

(i)

A = Event that three heads show up = {HHH}

B = Event that three tails show up = {TTT}

A $\cap$ B = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and B are mutually exclusive.

Question:5(ii) Three coins are tossed. Describe

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting no tails = {HHH}

B = Getting exactly one tail = {HHT, HTH, THH}

C = Getting at least two tails = {HTT, THT, TTH}

Clearly, A $\cap$ B = $\phi$ ; B $\cap$ C = $\phi$ ; C $\cap$ A = $\phi$

Since (A and B), (B and C) and (A and C) are mutually exclusive

Therefore A, B and C are mutually exclusive.

Also,

A $\cup$ B $\cup$ C = S

Hence A, B and C are exhaustive events.

Hence, A, B and C are three events which are mutually exclusive and exhaustive.

Question:5(iii). Three coins are tossed. Describe

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting at least one head = {HHH, HHT, HTH, THH, TTH}

B = Getting at most one head = {TTH, TTT}

Clearly, A $\cap$ B = {TTH} $\neq$ $\phi$

Hence, A and B are two events which are not mutually exclusive.

Question:5.(iv) Three coins are tossed. Describe

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting exactly one head = {HTT, THT, TTH}

B = Getting exactly one tail = {HHT, HTH, THH}

Clearly, A $\cap$ B = $\phi$

Hence, A and B are mutually exclusive.

Also, A $\cup$ B $\neq$ S

Hence, A and B are not exhaustive.

Question:5.(v) Three coins are tossed. Describe

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting exactly one tail = {HHT, HTH, THH}

B = Getting exactly two tails = {HTT, TTH, THT}

C = Getting exactly three tails = {TTT}

Clearly, A $\cap$ B = $\phi$ ; B $\cap$ C = $\phi$ ; C $\cap$ A = $\phi$

Since (A and B), (B and C) and (A and C) are mutually exclusive

Therefore A, B and C are mutually exclusive.

Also,

A $\cup$ B $\cup$ C = {HHT, HTH, THH, HTT, TTH, THT, TTT} $\neq$ S

Hence A, B and C are not exhaustive events.

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$ .

Describe the events

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

(i) Therefore, A'= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

= B : getting an odd number on the first die.

Question:6.(ii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$ .

Describe the events

not B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(ii) Therefore, B'= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

= A : getting an even number on the first die.

Question:âââââââ6.(iii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$ .

Describe the events

A or B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) A or B = A $\cup$ B = {(1,1), (1,2) .... (1,6), (3,1), (3,2).... (3,6), (5,1), (5,2)..... (5,6), (2,1), (2,2)..... (2,6), (4,1), (4,2)..... (4,6), (6,1), (6,2)..... (6,6)} = S

Question:6.(iv) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$

Describe the events

A and B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) A and B = A $\cap$ B = A $\cap$ A' = $\phi$ (From (ii))

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$

Describe the events

A but not C

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= { (2, 1), (2,2), (2,3) , (2,4), (2,5), (2,6), (4,1) , (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = { (1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(v) A but not C = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Question:.(vi) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$

Describe the events

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = { (1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1) }

(vi) B or C = B $\cup$ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3) , (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1) , (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

Question:6.(vii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$

Describe the events

B and C

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= { (1,1), (1,2), (1,3), (1,4) , (1,5), (1,6), (3,1), (3,2) , (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = { (1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(vi) B and C = B $\cap$ C = {(1, 1), (1,2), (1,3), (1,4), (3,1), (3,2)}

Question:6.(viii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice $\leq 5$

Describe the events

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= { (2, 1), (2,2), (2,3) , (2,4), (2,5), (2,6), (4,1) , (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3) , (3,1), (3,2), (4,1) }

(viii) A $\cap$ B' $\cap$ C' = A $\cap$ A $\cap$ C' (from (ii))

= A $\cap$ C' = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Here,

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) }

(i) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

A $\cap$ B = $\phi$ , since A and B have no common element amongst them.

Hence, A and B are mutually exclusive. TRUE

Here,

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) }

(ii) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

A $\cap$ B = $\phi$ , since A and B have no common element amongst them.

Hence, A and B are mutually exclusive.

Also,

A $\cup$ B = {(2,1), (2,2).... (2,6), (4,1), (4,2).....(4,6), (6,1), (6,2)..... (6,6), (1,1), (1,2).... (1,6), (3,1), (3,2)..... (3,6), (5,1), (5,2).... (5,6) } = S

Hence, A and B are exhaustive.

TRUE

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) Therefore, B' = S -B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} = A

TRUE

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = { (2, 1), (2,2), (2,3) , (2,4), (2,5), (2,6), (4,1) , (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3) , (3,1), (3,2), (4,1) }

(iv) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

A $\cap$ C = { (2,1), (2,2), (2,3) , (4,1) } ,

Hence, A and B are not mutually exclusive. FALSE

X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

A $\cap$ B' = A $\cap$ A = A (From (iii))

$\dpi{100} \therefore$ A $\cap$ B’ $\dpi{100} \neq \phi$

Hence A and B' not mutually exclusive. FALSE

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(vi) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\dpi{100} \therefore$ A' $\cap$ B' = B $\cap$ A = $\phi$ (from (iii) and (i))

Hence A' and B' are mutually exclusive.

Again,

$\dpi{100} \therefore$ B' $\cap$ C = A $\cap$ C $\dpi{80} \neq$ $\phi$ (from (iv))

Hence B' and C are not mutually exclusive.

Hence, A', B' and C are not mutually exclusive and exhaustive. FALSE

Class 11 maths chapter 16 NCERT solutions - Exercise: 16.3

 Assignment (a)

(a) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the assignment is valid

 Assignment (b)

(b) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = $\dpi{100} \frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7} = 1$

Therefore, the assignment is valid

 Assignment (c)

(c) Since sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1

Hence, Condition (ii) is not satisfied.

Therefore, the assignment is not valid

 Assignment (d)

(d) Two of the probabilities p( $\omega_1$ ) and p( $\omega_5$ ) are negative, hence condition(i) is not satisfied.

Therefore, the assignment is not valid.

 Assignment (e)

(e) Each of the number p( $\dpi{100} \omega_i$ ) is positive but p( $\dpi{100} \omega_7$ ) is not less than one. Hence the condition is not satisfied.

Therefore, the assignment is not valid.

Sample space when a coin is tossed twice, S = {HH, HT, TH, TT}

[Note: A coin tossed twice is same as two coins tossed at once]

$\therefore$ Number of possible outcomes n(S) = 4

Let E be the event of getting at least one tail = {HT, TH, TT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{4}$

= 0.75

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a prime number = {2,3,5}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{6}$

= 0.5

Question:3.(ii) A die is thrown, find the probability of following events:

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number greater than or equal to 3 = {3,4,5,6}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{6} = \frac{2}{3}$

= 0.67

Question:3.(iii) A die is thrown, find the probability of following events:

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than or equal to one = {1}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{6}$

= 0.167

Question:3.(iv) A die is thrown, find the probability of following events:

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number more than 6 will appear = $\phi$

$\therefore$ n(E) = 0

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{0}{6}$

= 0

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than 6 will appear = {1,2,3,4,5}

$\therefore$ n(E) = 5

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{5}{6}$

= 0.83

(a) Number of points(events) in the sample space = Number of cards in the pack = 52

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace of spades

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{52}$

The required probability that the card is an ace of spades is $\dpi{80} \frac{1}{52}$ .

Question:4(c)(i) A card is selected from a pack of 52 cards.

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace. There are 4 aces.

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{52} = \frac{1}{13}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{13}$ .

Number of possible outcomes, n(S) = 52

Let E be the event that the card is a black card. There are 26 black cards. (Diamonds and Clubs)

$\therefore$ n(E) = 26

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{26}{52} = \frac{1}{2}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{2}$ .

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of possible outcomes, n(S) = 12

(i) Let E be the event having sum of numbers as 3 = {(1, 2)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$

The required probability of having 3 as sum of numbers is $\dpi{80} \frac{1}{12}$ .

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of possible outcomes, n(S) = 12

(ii) Let E be the event having sum of numbers as 12 = {(6, 6)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$

The required probability of having 12 as sum of numbers is $\dpi{80} \frac{1}{12}$ .

There are four men and six women on the city council

$\therefore$ n(S) = n(men) + n(women) = 4 + 6 = 10

Let E be the event of selecting a woman

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{10} = \frac{3}{5}$

Therefore, the required probability of selecting a woman is 0.6

Here the sample space is,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}

According to question,

1.) 4 heads = 1 + 1 + 1 + 1 = Rs. 4

2.) 3 heads and 1 tail = 1 + 1 + 1 - 1.50 = Rs. 1.50

3.) 2 heads and 2 tails = 1 + 1 - 1.50 - 1.50 = - Rs. 1 : he will lose Re. 1

4.) 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 : he will lose Rs. 3.50

5.) 4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will lose Rs. 6

Now, sample space of amounts corresponding to S:

S' = {4, 1.50, 1.50, 1.50, 1.50, - 1, - 1, - 1, - 1, - 1, - 1, - 3.50, - 3.50, - 3.50, - 3.50, - 6}

$\therefore$ n(S') = 12

$\therefore$ Required Probabilities are:

$\dpi{100} P(Winning\ Rs.\ 4) = \frac{n(Winning\ Rs.\ 4)}{n(S')}$ $= \frac{1}{16}$

$\dpi{100} P(Winning\ Rs.\ 1.50) = \frac{n(Winning\ Rs.\ 1.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Re.\ 1) = \frac{n(Losing\ Re.\ 1)}{n(S')}$ $= \frac{6}{16} = \frac{3}{8}$

$\dpi{100} P(Losing\ Rs.\ 3.50) = \frac{n(Losing\ Rs.\ 3.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Rs.\ 6) = \frac{n(Losing\ Rs.\ 6)}{n(S')}$ $= \frac{1}{16}$

$\small 3$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 3 heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting 3 heads is $\dpi{80} \frac{1}{8}$ .

Question 8.(ii) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 2 heads = {HHT, HTH, THH}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$

The required probability of getting 2 heads is $\dpi{80} \frac{3}{8}$ .

Question:âââââââ8.(iii) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atleast 2 heads = Event of getting 2 or more heads = {HHH, HHT, HTH, THH}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{8} = \frac{1}{2}$

The required probability of getting atleast 2 heads is $\dpi{80} \frac{1}{2}$ .

Question:8.(iv) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 heads = Event of getting 2 or less heads = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$= \frac{6}{8} = \frac{3}{4}$ $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$

The required probability of getting almost 2 heads is $\dpi{80} \frac{3}{4}$ .

Question:8.(v) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting no head = Event of getting only tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting no head is $\dpi{80} \frac{1}{8}$ .

Question:8.(vi) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 3 tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting 3 tails is $\dpi{80} \frac{1}{8}$ .

Question:8(vii) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$

The required probability of getting exactly 2 tails is $\dpi{80} \frac{3}{8}$ .

Question:8.(viii) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting no tail = Event of getting only heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting no tail is $\dpi{80} \frac{1}{8}$ .

Question:8.(ix) Three coins are tossed once. Find the probability of getting

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{8} = \frac{3}{4}$

The required probability of getting atmost 2 tails is $\dpi{80} \frac{3}{4}$ .

Given,

P(E) = $\small \frac{2}{11}$

We know,

P(not E) = P(E') = 1 - P(E)

= $\dpi{80} 1 - \small \frac{2}{11}$

= $\dpi{80} \frac{9}{11}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of vowels = {3 A's,2 I's,O} = 6

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a vowel.

n(E) = $\dpi{100} ^{6}\textrm{C}_{1}$ = 6

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{6}{13}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of consonants = {4 S's,2 N's,T} = 7

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a consonant.

n(E) = $\dpi{100} ^{7}\textrm{C}_{1}$ = 7

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{7}{13}$

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

$\dpi{100} \therefore$ n(S) = $\dpi{100} ^{20}\textrm{C}_{6}$

Let E be the event that six numbers match with the six numbers fixed by the lottery committee.

n(E) = 1 (Since only one prize to be won.)

$\dpi{100} \therefore$ Probability of winning =

$\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $\dpi{100} = \frac{1}{^{20}\textrm{C}_{6}} = \frac{6!14!}{20!}$

$\dpi{100} = \frac{6.5.4.3.2.1.14!}{20.19.18.17.16.15.14!}$

$\dpi{100} = \frac{1}{38760}$

(i) Given, $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

Now P(A $\cap$ B) > P(A)

(Since A $\cap$ B is a subset of A, P( A $\cap$ B) cannot be more than P(A))

Therefore, the given probabilities are not consistently defined.

(ii) Given, $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.9 - 0.8 = 0.1

Therefore, P(A $\cap$ B) < P(A) and P(A $\cap$ B) < P(B) , which satisfies the condition.

Hence, the probabilities are consistently defined

Question:13 Fill in the blanks in following table:

 (i) (ii) (iii)

We know,

$\dpi{100} P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

(i) $\dpi{100} P(A \cup B)$ = $\dpi{100} \frac{1}{3}+\frac{1}{5}-\frac{1}{15}$ = $\dpi{100} \frac{5+3-1}{15} = \frac{7}{15}$

(ii) $\dpi{100} 0.6 = 0.35 + P(B) - 0.25$

$\implies$ $\dpi{100} P(B) = 0.6 - 0.1 = 0.5$

(iii) $\dpi{100} 0.7 = 0.5 + 0.35 - P(A \cap B)$

$\implies$ $\dpi{100} P(A \cap B) = 0.85 - 0.7 = 0.15$

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $\dpi{100} \boldsymbol{\frac{7}{15}}$ (ii) $0.35$ 0.5 $0.25$ $0.6$ (iii) $0.5$ $0.35$ 0.15 $0.7$

Given, $\dpi{80} P(A)=\frac{3}{5}$ and $\dpi{80} P(B)=\frac{1}{5}$

To find : $\dpi{100} P(A or B) = P(A \cup B)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B) = P(A)+ P(B)$ [Since A and B are mutually exclusive events.]

$\implies$ $P(A \cup B) = \frac{3}{5}+\frac{1}{5} = \frac{4}{5}$

Therefore, $\dpi{100} P(A \cup B) = \frac{4}{5}$

Given, $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(E or F) = P(E \cup F)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$
Therefore, $P(E \cup F) =$ $\frac{5}{8}$

Given, $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(not\ E\ and\ not\ F) = P(E' \cap F')$

We know,

$P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$

And $P(A\cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$

$\implies$ $P(E' \cap F') = 1 - P(E \cup F)$

$= 1- \frac{5}{8}= \frac{3}{8}$

Therefore, $P(E' \cap F') =$ $\frac{3}{8}$

Given, $P(not\ E\ or\ not\ F) = 0.25$

For A and B to be mutually exclusive, $P(A \cap B) = 0$

Now, $P(not\ E\ or\ not\ F) = P(E' \cup F') = 0.25$

We know,

$P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B)$
$\\ \implies 0.25 = 1 - P(E \cap F) \\ \implies P(E \cap F) = 1 - 0.25 = 0.75 \neq 0$

Hence, E and F are not mutually exclusive.

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(i) $P(not\ A) = P(A') = 1 - P(A)$

$\implies$ $P(not\ A) = 1 - 0.42 = 0.58$

Therefore, P(not A) = 0.58

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(ii) $P(not\ B) = P(B') = 1 - P(B)$

$\implies$ $P(not\ B) = 1 - 0.48 = 0.52$

Therefore, P(not B) = 0.52

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(iii) We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.9 - 0.16$

= 0.74

Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology

And total students in the class be 100.

Given, n(M) = 40 $\implies$ P(M) = $\dpi{100} \frac{40}{100} = \frac{2}{5}$

n(B) = 30 $\implies$ P(M) = $\dpi{100} \frac{30}{100} = \frac{3}{10}$

n(M $\cap$ B) = 10 $\implies$ P(M) = $\dpi{100} \frac{10}{100} = \frac{1}{10}$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(M $\cup$ B) = 0.4 + 0.3 - 0.1 = 0.6

Hence, the probability that he will be studying Mathematics or Biology is 0.6

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination.

P(A $\cup$ B) is probability of passing at least one of the examination.

Therefore,

P(A $\cup$ B) = 0.95 , P(A)=0.8, P(B)=0.7

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55

Hence,the probability that the student will pass both the examinations is 0.55

Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination.

Given,

P(A)=0.75, P(A $\cap$ B) = 0.5, P(A' $\cap$ B') =0.1

We know,

P(A' $\cap$ B') = 1 - P(A $\cup$ B)

$\implies$ P(A $\cup$ B) = 1 - 0.1 = 0.9

Also,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(B) = 0.9 - 0.75 + 0.5 = 0.65

Hence,the probability of passing the Hindi examination is 0.65

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(i) We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\frac{1}{2} + \frac{8}{15} - \frac{2}{5} = \frac{15+16-12}{30}$

$= \frac{19}{30}$

Hence,the probability that the student opted for NCC or NSS is $\frac{19}{30}$

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(ii) Now,

Probability that the student has opted neither NCC nor NSS = P(A' $\cap$ B' )

We know,

P(A' $\cap$ B' ) = 1 - P(A $\cup$ B) [De morgan's law]

And, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$= \frac{19}{30}$

$\therefore$ P(A' $\cap$ B' )

$= 1- \frac{19}{30} = \frac{11}{30}$

Hence,the probability that the student has opted neither NCC nor NSS. is $\frac{11}{30}$

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(iii) Now,

Probability that the student has opted NSS but not NCC = P(B $\cap$ A' ) = P(B-A)

We know,

P(B-A) = P(B) - P(A $\cap$ B)

$= \frac{8}{15}- \frac{2}{5} = \frac{8-6}{15}$

$= \frac{2}{15}$

Hence,the probability that the student has opted NSS but not NCC is $\frac{2}{15}$

Class 11 Probability NCERT solutions - Miscellaneous Exercise

Given,

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = $^{60}\textrm{C}_{5}$

(i) .

Number of ways of drawing 5 blue marbles from 20 blue marbles = $^{20}\textrm{C}_{5}$

$\therefore$ Probability of drawing all blue marbles = $\frac{^{20}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

Given,

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = $^{60}\textrm{C}_{5}$

(ii). We know,

The probability that at least one marble is green = 1 - Probability that no marble is green

Now, Number of ways of drawing no green marbles = Number of ways of drawing only red and blue marbles

= $^{(20+10)}\textrm{C}_{5} = ^{30}\textrm{C}_{5}$

$\therefore$ The probability that no marble is green = $\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

$\therefore$ The probability that at least one marble is green = $1-\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

Total number of ways of drawing 4 cards from a deck of 52 cards = $^{52}\textrm{C}_{4}$

We know that there are 13 diamonds and 13 spades in a deck.

Now, Number of ways of drawing 3 diamonds and 1 spade = $^{13}\textrm{C}_{3}.^{13}\textrm{C}_{1}$

$\therefore$ Probability of obtaining 3 diamonds and 1 spade

$= \frac{^{13}\textrm{C}_{3}.^{13}\textrm{C}_{1}}{^{52}\textrm{C}_{4}}$

Total number of faces of a die = 6

(i) Number of faces with number '2' = 3

$P(2) = \frac{3}{6}$ $=\frac{1}{2}$

Therefore, required probability P(2) is 0.5

Total number of faces of a die = 6

(ii) P (1 or 3) = P (not 2) = 1 − P (2)

$= 1-P(2) = 1-\frac{3}{6}$ $=\frac{1}{2}$

Therefore, required probability P(1 or 3) is 0.5

Total number of faces of a die = 6

(iii) Number of faces with number '3' = 1

$\therefore$ $P(3) = \frac{1}{6}$

$\therefore$ P(not 3) = 1 - P(3)

$= 1-\frac{1}{6} = \frac{5}{6}$

Therefore, required probability P(not 3) is $\frac{5}{6}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(a) If one ticket is bought,

P(getting a prize) = $\frac{10}{10000} = \frac{1}{1000}$

$\therefore$ P(not getting a prize) = 1 - P(getting a prize)

$1- \frac{1}{1000} = \frac{999}{1000}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(b) If two tickets are bought:

Number of tickets not awarded = 10000 - 10 = 9990

$\therefore$ P(not getting a prize) = $\frac{^{9990}\textrm{C}_{2}}{^{10000}\textrm{C}_{2}}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(c) If ten tickets are bought:

Number of tickets not awarded = 10000 - 10 = 9990

$\therefore$ P(not getting a prize) = $\frac{^{9990}\textrm{C}_{10}}{^{10000}\textrm{C}_{10}}$

you both enter the same section?

Total number of students = 100

Let A and B be the two sections consisting of 40 and 60 students respectively.

Number of ways of selecting 2 students out of 100 students.= $^{100}\textrm{C}_{n}$

If both are in section A:

Number of ways of selecting 40 students out of 100 = $\dpi{100} ^{100}\textrm{C}_{40}$ (The remaining 60 will automatically be in section B!)

Remaining 38 students are to be chosen out of (100-2 =) 98 students

$\dpi{100} \therefore$ Required probability if they both are in section A = $\frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}$

Similarly,

If both are in section B:

Number of ways of selecting 60 students out of 100 = $\dpi{100} = ^{100}\textrm{C}_{60} = ^{100}\textrm{C}_{40}$ (The remaining 40 will automatically be in section A!)

Remaining 58 students are to be chosen out of (100-2 =) 98 students

$\dpi{100} \therefore$ Required probability if they both are in section B = $\frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}$

Required probability that both are in same section = Probability that both are in section A + Probability that both are in section B

= $\frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}+\frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}$

$\\ = \frac{^{98}\textrm{C}_{38}+^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{40}} \\ \\ =\frac{\frac{98!}{38!.60!} + \frac{98!}{58!.40!}}{\frac{100!}{40!.60!}}$

$=\frac{85}{165} = \frac{17}{33}$

Hence, the required probability that both are in same section is $\dpi{100} \frac{17}{33}$

you both enter the different sections?

Total number of students = 100

Let A and B be the two sections consisting of 40 and 60 students respectively.

We found out in (a) that the probability that both students are in same section is $\dpi{100} \frac{17}{33}$

(b) Probability that both the students are in different section = $\dpi{100} 1 - \frac{17}{33}$ $\dpi{100} = \frac{16}{33}$

Given, 3 letters are put in 3 envelopes.

The number of ways of putting the 3 different letters randomly = 3!

Number of ways that at least one of the 3 letters is in the correct envelope

= No. of ways that exactly 1 letter is in correct envelope + No. of ways that 2 letters are in the correct envelope(The third is automatically placed correctly)

= No. of ways that exactly 1 letter is in correct envelope + No. of ways that all the 3 letters are in the correct envelope

= $(^{3}\textrm{C}_{1}\times1) + 1 = 4$

(Explanation for $^{3}\textrm{C}_{1}\times1$ :

No. of ways of selecting 1 envelope out of 3 = $^{3}\textrm{C}_{1}$ .

If we put the correct letter in it, there is only one way the other two are put in the wrong envelope! )

Therefore, the probability that at least one letter is in its proper envelope = $\frac{4}{3!} = \frac{4}{6} = \frac{2}{3}$

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

(i) We know, P(A $\dpi{100} \cup$ B) = P(A) + P(B) - P(A $\dpi{100} \cap$ B)

$\dpi{100} \implies$ P(A $\dpi{100} \cup$ B) = 0.54 + 0.69 - 0.35 = 0.88

$\dpi{100} \implies$ P(A $\dpi{100} \cup$ B) = 0.88

Given, $P(A) = 0.54, P(B) = 0.69, P(A \cap B) = 0.35$

And, $P(A \cup B) = 0.88$

(ii) $A' \cap B'= (A \cup B)'$ [De Morgan’s law]

So, $P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B) = 1 - 0.88 = 0.12$

$\therefore$ $P(A' \cap B') = 0.12$

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

And, P(A $\dpi{100} \cup$ B) = 0.88

(iii) P(A $\dpi{100} \cap$ B') = P(A-B) = P(A) - P(A $\dpi{100} \cap$ B)

= 0.54 - 0.35 = 0.19

$\therefore$ P(A $\dpi{100} \cap$ B') = 0.19

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

And, P(A $\dpi{100} \cup$ B) = 0.88

(iv) P(B $\dpi{100} \cap$ A') = P(B-A) = P(B) - P(A $\dpi{100} \cap$ B)

= 0.69 - 0.35 = 0.34

$\therefore$ P(B $\dpi{100} \cap$ A') = 0.34

 S. No. Name Sex Age in years 1. Harish M 30 2. Rohan M 33 3. Sheetal F 46 4. Alis F 28 5. Salim M 41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $\small 35$ years?

Given,

Total number of persons = 5

No. of male spokesperson = 3

No. of spokesperson who is over 35 years of age = 2

Let E be the event that the spokesperson is a male and F be the event that the spokesperson is over 35 years of age.

$\therefore$ $P(E) = \frac{3}{5}\ and\ P(F) = \frac{2}{5}$

Since only one male is over 35 years of age,

$\therefore$ $P(E\cap F) = \frac{1}{5}$

We know,

$P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$\implies P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5}$

Therefore, the probability that the spokesperson will either be a male or over 35 years of age is $\frac{4}{5}$ .

Since 4-digit numbers greater than 5000 are to be formed,

The $1000's$ place digit can be filled up by either 7 or 5 in $^{2}\textrm{C}_{1}$ ways

Since repetition is allowed,

Each of the remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 in $5$ ways.

$\therefore$ Total number of 4-digit numbers greater than 5000 = $^{2}\textrm{C}_{1}\times5\times5\times5 -1$

$= 250 - 1 = 249$ (5000 cannot be counted, hence one less)

We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = $^{2}\textrm{C}_{1}\times5\times5\times^{2}\textrm{C}_{1} -1$ $= 100 - 1 = 99$

Therefore, the required probability =

$P(with\ repetition) = \frac{99}{249} = \frac{33}{83}$

Since 4-digit numbers greater than 5000 are to be formed,

The $1000's$ place digit can be filled up by either 7 or 5 in $^{2}\textrm{C}_{1}$ ways

Since repetition is not allowed,

The remaining 3 places can be filled by remaining 4 digits in $^{4}\textrm{C}_{3}\times3!$ ways.

$\therefore$ Total number of 4-digit numbers greater than 5000 = $^{2}\textrm{C}_{1}\times(^{4}\textrm{C}_{3}\times3!)= 2\times4\times6 = 48$

We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.

Case 1. When digit at $1000's$ place is 5, the units place can be filled only with 0.

And the $100's$ & $10's$ places can be filled with any two of the remaining digits {1,3,7} in

$\therefore$ Number of 4-digit numbers starting with 5 and divisible by 5 =

Case 2. When digit at $1000's$ place is 7, the units place can be filled by 0 or 5 in 2 ways.

And the $100's$ & $10's$ places can be filled with any two of the remaining 3 digits in

$\therefore$ Number of 4-digit numbers starting with 7 and divisible by 5 =

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Therefore, the required probability =

$P(without\ repetition) = \frac{18}{48} = \frac{3}{8}$

Given, Each wheel can be labelled with 10 digits.

Number of ways of selecting 4 different digits out of the 10 digits = $^{10}\textrm{C}_{4}$

These 4 digits can arranged among themselves is $4!$ ways.

$\therefore$ Number of four digit numbers without repetitions =

$^{10}\textrm{C}_{4}\times4! = \frac{10!}{4!.6!}\times4! = 10\times9\times8\times7 = 5040$

Number of combination that can open the suitcase = 1

$\therefore$ Required probability of getting the right sequence to open the suitcase = $\frac{1}{5040}$

## Summary Of NCERT Solutions For Class 11 Maths Chapter 16 Probability

Section 16.1 - Introduction

This section presents a concise explanation of the concept of probability and its associated terminology. In basic terms, probability refers to the proportion of favorable outcomes to the total number of possible outcomes. Other relevant terms such as experiment, sample space, events, and outcomes are also discussed.

Section 16.2 – Random Experiments

This section, 16.2.1, focuses on outcomes and sample space in random experiments. The definition of a random experiment, along with the necessary conditions and relevant terms, is elaborated upon with the help of examples such as coin tossing and rolling dice. This approach aims to facilitate a clear understanding of the topic for students.

Section 16.3 – Event

The following sub-topics have been covered in this section, 16.3:

• 16.3.1: The probability of an event occurring
• 16.3.2: Different types of events
• 16.3.3: The algebra of events, which covers concepts like intersection, union, difference, and complement of events.
• 16.3.4: Mutually exclusive events
• 16.3.5: Exhaustive events

Numerous examples and experiments have been used to elucidate these topics and provide a clear understanding of the related terms and event types. The algebra of events is particularly helpful in comprehending concepts such as intersection, union, difference, and complement of events.

Section 16.4 – Axiomatic Approach to Probability

This section, 16.4, covers various methods of calculating probabilities. The sub-topics are as follows:

• 16.4.1: Probability of an event
• 16.4.2: Probabilities of equally likely outcomes
• 16.4.3: Probability of the event 'A or B'
• 16.4.4: Probability of the event 'not A'

The axiomatic approach is another way to define probability, and this is comprehensively explained in Class 11 Maths Chapter 16 of the NCERT curriculum. Furthermore, numerous examples and practice problems are available to aid in the comprehension of these concepts.

Intrested students can practice class 11 maths ch 16 question answer using the following exercises.

## NCERT Solutions for Class 11 Mathematics

 chapter-1 Sets chapter-2 Relations and Functions chapter-3 Trigonometric Functions chapter-4 Principle of Mathematical Induction chapter-5 Complex Numbers and Quadratic equations chapter-6 Linear Inequalities chapter-7 Permutation and Combinations chapter-8 Binomial Theorem chapter-9 Sequences and Series chapter-10 Straight Lines chapter-11 Conic Section chapter-12 Three Dimensional Geometry chapter-13 Limits and Derivatives chapter-14 Mathematical Reasoning chapter-15 Statistics chapter-16 Probability

## Key Features Of NCERT Solutions For Maths Chapter 16 class 11 Probability

Some key features of ncert solutions for class 11 maths chapter 16 :

Comprehensive Coverage: The probability class 11 solutions provide a detailed explanation of all the key concepts of probability, including sample space, events, probability distribution, conditional probability, and more.

Easy-to-understand language: The ch 16 maths class 11 solutions are written in simple language, making it easy for students to comprehend and learn the concepts without any difficulty.

Chapter-wise Exercise Solutions: The class 11 probability NCERT solutions provide chapter-wise exercise solutions, which include both short answer and long answer type questions, along with step-by-step solutions.

## NCERT Books and NCERT Syllabus

1. What are important topics of the probability class 11 ncert solutions ?

The ch 16 maths class 11  includes topics such as Basic probability problems, Random Experiments, Events, and Axiomatic Approach to Probability. students can practice these topics from NCERT solutions as discussed above in this article.

2. In chapter 16 class 11th maths, what is meant by Conditional Probability?

The concept of Conditional Probability in chapter 16 class 11 maths refers to the situation where the occurrence of one event is dependent on the occurrence of another event. In this case, both events are considered to be independent of each other, meaning that the probability of event A does not affect the probability of event B. The solutions provide a detailed explanation of this concept, which can help students score well in their Class 11 exam.

3. In Chapter 16 of the NCERT Solutions for Class 11 Maths, can you provide an explanation of the concept of sample space?

In the NCERT Solutions for Class 11 Maths Chapter 16, the concept of sample space refers to a set of all possible outcomes of a future event, such as picking a card from a deck of cards. This set is defined by the method used to select an item from a given sample, and every probability is a subset of the sample space. It is important to note that the sum of all probabilities within a sample space is always equal to one. The solutions for this chapter, available in PDF format, provide a detailed explanation of this concept to help students better understand it.

4. List out the major topics and subtopics covered in Chapter 16class 11 maths probability.

there are major topics and sub topics in class 11th maths chapter 16:

• Introduction (16.1)
• Random Experiments (16.2)
• Outcomes and Sample Space (16.2.1)
• Event (16.3)
• Occurrence of an Event (16.3.1)
• Types of Events (16.3.2)
• Algebra of Events (16.3.3)
• Mutually Exclusive Events (16.3.4)
• Exhaustive Events (16.3.5)
• Axiomatic Approach to Probability (16.4)
• Probability of an Event (16.4.1)
• Probabilities of Equally Likely Outcomes (16.4.2)
• Probability of the Event 'A or B' (16.4.3)
• Probability of Event 'Not A' (16.4.4)
5. Where can I find the complete solutions of class 11th maths probability ?

Here you will get the detailed NCERT solutions for class 11 maths  by clicking on the link. students can these solutions to command and indepth understanding the concepts. this will help students to build confidence overall lead to score well in the exams. for ease they can study probability class 11 pdf both online and offline mode.

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Get answers from students and experts

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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