NCERT Solutions for Class 11 Maths Chapter 14 Probability

Question:1. A die is rolled. Let E be the event “die shows $4$ ” and F be the event “die shows even number”. Are E and F mutually exclusive?

Answer:

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 $\cap$ F = {4} $\cap$ {2, 4, 6}

= {4} $\ne \varphi$

Hence E and F are not mutually exclusive events.

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

(i) A: a number less than 7 (ii) B: a number greater than 7
(iii) C: a multiple of 3. (iv) D: a number less than 4
(v) E: an even multiple greater than 4 (vi) F: a number not less than 3

Also, find $A\cup B$, $A\cap B$, $B\cup C$, $E\cap F$, $D\cap E$, $A-C$, $D-E$, $E\cap F^{\prime }$, $F^{\prime }$

Answer:

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

S = {1, 2, 3, 4, 5, 6} or {x: x $\in$ 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

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

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

Given, B: a number greater than 7

As no number on the die is greater than 7

B = $\varphi$

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

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

Given, C: a multiple of 3

C = {3, 6}

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

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

Given, D: a number less than 4

D = {1, 2, 3}

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

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

Given, E: an even number greater than 4

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

Therefore , E = {6}

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

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

Given, F: a number not less than 3

F = {x: x $\in$ S, x $\ge$ 3 } = {3, 4, 5, 6}

Now,

A $\cup$ B = { 1, 2, 3, 4, 5, 6 } $\cup$$\varphi$ = {1, 2, 3, 4, 5, 6}

A $\cap$ B = {1, 2, 3, 4, 5, 6} $\cap$$\varphi$ = $\varphi$

B $\cup$ C = $\varphi$$\cup$ { 3, 6 } = {3, 6}

E $\cap$ F = { 6 } $\cap$ {3, 4, 5, 6 } = {6}

D $\cap$ E = {1, 2, 3} $\cap$ {6} = $\varphi$ (As nothing is common in these sets)

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

D - E = {1, 2, 3} - {6} = {1, 2, 3}

F' = {3, 4, 5, 6}' = S - F = {1, 2}

$\therefore$ E $\cap$ F' = {6} $\cap$ {1, 2} = $\varphi$

Question:3 An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

(a) the sum is greater than $8$

(b) $2$ occurs on either die

(c) the sum is at least $7$ and a multiple of $3$

(d) Which pairs of these events are mutually exclusive?

Answer:

(a) 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 $\in$ 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 are9, 10, 11, and 12

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

(b) 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 $\in$ 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 the first die or second die or on both die simultaneously.

B = [ {(a,b): (a,b) $\in$ 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)}

(c) 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 $\in$ 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) $\in$ E, a+b>6 & a+b = 3k, k $\in$ I} ] = {(3,6), (6,3), (5, 4), (4,5), (6,6)}

(d) 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 = $\varphi$ (no common element in A and B)

Hence, A and B are mutually exclusive

Again, B $\cap$ C = $\varphi$ (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.

Question:4 Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event a head shows on the first coin”. Which events are

(i) mutually exclusive?

(ii) simple?

(iii) Compound?

Answer:

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 = $\varphi$

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

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

C $\cap$ D = {TTT} $\cap$ {HHH, HHT, HTH, HTT} = $\varphi$ ; 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} = $\varphi$ ; 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.

(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.

(iii). 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 Three coins are tossed. Describe

(i) Two mutually exclusive events.

(ii) Three events which are mutually exclusive and exhaustive.

(iii) Two events, which are not mutually exclusive

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive

Answer:

(i) 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} = $\varphi$ ; Hence A and B are mutually exclusive.

(ii) 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 = $\varphi$ ; B $\cap$ C = $\varphi$ ; C $\cap$ A = $\varphi$

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 that are mutually exclusive and exhaustive.

(iii) 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} $\ne$ $\varphi$

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

(iv) 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 = $\varphi$

Hence, A and B are mutually exclusive.

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

Hence, A and B are not exhaustive.

(v) 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 = $\varphi$ ; B $\cap$ C = $\varphi$ ; C $\cap$ A = $\varphi$

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} $\ne$ S

Hence A, B, and C are not exhaustive events.

Question:6.(i) 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 $\le 5$.

Describe the events

(i) $A{}^{\prime }$

(ii) not B

(iii) A or B

(iv) A and B

(v) A but not C

(vi) B or C

(vii) B and C

(viii) $A\cap B^{\prime }\cap C^{\prime }$

Answer:

(i) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

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

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.

(ii) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

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

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.

(iii) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

A: getting an even number on the first die = {(a,b): a $\in$ {2,4,6} and 1 $\le$ b $\le$ 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 $\in$ {1,3,5} and 1 $\le$ b $\le$ 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)}

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

(iv) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

A: getting an even number on the first die = {(a,b): a $\in$ {2,4,6} and 1 $\le$ b $\le$ 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 $\in$ {1,3,5} and 1 $\le$ b $\le$ 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)}

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

(v) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

A: getting an even number on the first die = {(a,b): a $\in$ {2,4,6} and 1 $\le$ b $\le$ 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 $\le$ 5

The possible sum is 2,3,4,5

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

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)}

(vi) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

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

The possible sum is 2,3,4,5

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

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)}

(vii) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

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

The possible sum ais2,3,4,5

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

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

(viii) Sample space when two dice are thrown:

S = {(x,y): 1 $\le$ x,y $\le$ 6}

A: getting an even number on the first die = {(a,b): a $\in$ {2,4,6} and 1 $\le$ b $\le$ 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 $\in$ {1,3,5} and 1 $\le$ b $\le$ 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 $\le$ 5

The possible sum aris,3,4,5

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

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)}

Question:7.(i) Refer to question 6 above, state true or false: (give a reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii) $A=B{}^{\prime }$

(iv) A and C are mutually exclusive

(v) $A$ and $B^{\prime }$ are mutually exclusive

(vi) $A^{\prime },B^{\prime },C$ are mutually exclusive and exhaustive.

Answer:

(i) 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 = $\varphi$

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

Hence, A and B are mutually exclusive. TRUE

(ii) 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 = $\varphi$

A $\cap$ B = $\varphi$ , 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

(iii) Here,

S = {(x,y): 1 $\le$ x,y $\le$ 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

(iv) Here,

S = {(x,y): 1 $\le$ x,y $\le$ 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 = $\varphi$

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

Hence, A and B are not mutually exclusive. FALSE

(v) X and Y are mutually exclusive if and only if X $\cap$ Y = $\varphi$

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

$\therefore$ A $\cap$ B’ $\ne \varphi$

Hence A and B' are not mutually exclusive. FALSE

(vi) Here,

S = {(x,y): 1 $\le$ x,y $\le$ 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 = $\varphi$

$\therefore$ A' $\cap$ B' = B $\cap$ A = $\varphi$ (from (iii) and (i))

Hence A' and B' are mutually exclusive.

Again,

$\therefore$ B' $\cap$ C = A $\cap$ C $\ne$ $\varphi$ (from (iv))

Hence B' and C are not mutually exclusive.

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

Question:1(a) Which of the following can not be valid assignment of probabilities for outcomes of sample Space S $=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(a) Condition (i): Each of the numbers p( ${\omega }_i$ ) 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.

Question:1.(b) Which of the following cannot be the valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(b) Condition (i): Each of the numbers p( ${\omega }_i$ ) is positive and less than one.

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

Therefore, the assignment is valid.

Question:1.(c) Which of the following can not be a valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(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.

Question:1.(d) Which of the following can not be a valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(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.

Question:1.(e) Which of the following can not be a valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(e) Each of the number p( ${\omega }_i$ ) is positive but p( ${\omega }_7$ ) is not less than one. Hence, the condition is not satisfied.

Therefore, the assignment is not valid.

Question:2 A coin is tossed twice, what is the probability that at least one tail occurs?

Answer:

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

[Note: A coin tossed twice is the 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

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

= 0.75

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

(i) A prime number will appear

(ii) A number greater than or equal to $3$ will appear

(iii) A number less than or equal to one will appear

(iv) A number more than $6$ will appear

(v) A number less than $6$ will appear.

Answer:

(i) 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

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

= 0.5

(ii) 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

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

= 0.67

(iii) 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

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

= 0.167

(iv) 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 = $\varphi$

$\therefore$ n(E) = 0

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

= 0

(v) 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

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

= 0.83

Question:4. A card is selected from a pack of $52$ cards.
(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c)(i) Calculate the probability that the card is an ace.

(c)(ii) Calculate the probability that the card is a black card.

Answer:

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

(b) Number of possible outcomes, n(S) = 52

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

$\therefore$ n(E) = 1

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

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

(c)(i) 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

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

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

(c)(ii) Number of possible outcomes, n(S) = 52

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

$\therefore$ n(E) = 26

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

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

Question:5. A fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is

(i) $3$

(ii) $12$

Answer:

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 possible outcomes

Sample space, S = {(x,y): x $\in$ {1,6} and y $\in$ {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 the sum of numbers as 3 = {(1, 2)}

$\therefore$ n(E) = 1

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

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

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

$\therefore$ n(E) = 1

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

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

Question:6 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Answer:

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

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

Therefore, the required probability of selecting a woman is 0.6

Question:7. A fair coin is tossed four times, and a person wins Rs $1$ for each head and lose Rs $1.50$ for each tail that turns up. From the sample, space calculates how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Answer:

Here the sample space is,

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

According to the 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, the 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:

$P(WinningRs.4)=\frac{n(WinningRs.4)}{n(S^{\prime })}$ $=\frac{1}{16}$

$P(WinningRs.1.50)=\frac{n(WinningRs.1.50)}{n(S^{\prime })}$ $=\frac{4}{16}=\frac{1}{4}$

$P(LosingRe.1)=\frac{n(LosingRs.1)}{n(S^{\prime })}$ $=\frac{6}{16}=\frac{3}{8}$

$P(LosingRs.3.50)=\frac{n(LosingRs.3.50)}{n(S^{\prime })}$ $=\frac{4}{16}=\frac{1}{4}$

$P(LosingRs.6)=\frac{n(LosingRs.6)}{n(S^{\prime })}$ $=\frac{1}{16}$

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

(i) $3$ heads

(ii) $2$ heads

(iii) atleast $2$ heads

(iv) atmost $2$ heads

(v) no head

(vi) $3$ tails

(vii) exactly two tails

(viii) no tail

(ix) atmost two tails

Answer:

(i) 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

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

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

(ii) 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

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

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

(iii) 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

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

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

(iv) 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

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

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

(v) 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

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

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

(vi) 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

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

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

(vii) 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

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

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

(viii) 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

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

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

(ix) 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

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

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

Question:9 If $\frac{2}{11}$ is the probability of an event, what is the probability of the event ‘not A’?

Answer:

Given,

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

We know,

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

= $1-\frac{2}{11}$

= $\frac{9}{11}$

Question: 10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that

(i) letter is a vowel

(ii) that letter is a consonant

Answer:

(i) 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) = ${}^{13}{\text{C}}_1$ = 13

Let E be the event of getting a vowel.

n(E) = ${}^6{\text{C}}_1$ = 6

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

(ii) 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) = ${}^{13}{\text{C}}_1$ = 13

Let E be the event of getting a consonant.

n(E) = ${}^7{\text{C}}_1$ = 7

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

Question:11 In a lottery, a person chooses six different natural numbers at random from $1$ to $20$ , and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [ Hint order of the numbers is not important.]

Answer:

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

$\therefore$ n(S) = ${}^{20}{\text{C}}_6$

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.)

$\therefore$ Probability of winning =

$P(E)=\frac{n(E)}{n(S)}$ $=\frac{1}{{}^{20}{\text{C}}_6}=\frac{6!14!}{20!}$

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

$=\frac{1}{38760}$

Question:12. Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined:

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

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

Answer:

(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)

$⟹$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B)

$⟹$ 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 the following table:

Answer:

We know,

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

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

(ii) $0.6=0.35+P(B)-0.25$

$⟹$ $P(B)=0.6-0.1=0.5$

(iii) $0.7=0.5+0.35-P(A\cap B)$

$⟹$ $P(A\cap B)=0.85-0.7=0.15$

Question:14 Given $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$ . Find $P(AorB)$ , if $A$ and $B$ are mutually exclusive events.

Answer:

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

To find : $P(AorB)=P(A\cup B)$

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.]

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

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

Question:15 If E and F are events such that $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(EandF)=\frac{1}{8}$ , find

(i) P(E or F)

(ii) P(not E and not F)

Answer:

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

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

We know,

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

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

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

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

To find : $P(notEandnotF)=P(E^{\prime }\cap F^{\prime })$

We know,

$P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }=1-P(A\cup B)$

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

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

$=\frac{5}{8}$

$⟹$ $P(E^{\prime }\cap F^{\prime })=1-P(E\cup F)$

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

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

Question:16 Events E and F are such that P(not E or not F) $=0.25$ , State whether E and F are mutually exclusive.

Answer:

Given, $P(notEornotF)=0.25$

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

Now, $P(notEornotF)=P(E^{\prime }\cup F^{\prime })=0.25$

We know,

$P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }=1-P(A\cap B)$
$$\begin{gathered} ⟹0.25=1-P(E\cap F) \\ ⟹P(E\cap F)=1-0.25=0.75\ne 0 \end{gathered}$$

Hence, E and F are not mutually exclusive.

Question:17 A and B are events such that P(A) $=0.42$, P(B) $=0.48$ and P(A and B) . $=0.16$ Determine

(i) (P(not A)

(ii) P(not B)

(iii) P(A or B)

Answer:

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

(i) $P(notA)=P(A^{\prime })=1-P(A)$

$⟹$ $P(notA)=1-0.42=0.58$

Therefore, P(not A) = 0.58

(ii) $P(notB)=P(B^{\prime })=1-P(B)$

$⟹$ $P(notB)=1-0.48=0.52$

Therefore, P(not B) = 0.52

(iii) We know,

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

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

= 0.74

Question:18 In Class XI of a school $40\mathrm{\%}$ of the students study Mathematics and $30\mathrm{\%}$ study Biology. $10\mathrm{\%}$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Answer:

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

And the total number of students in the class is 100.

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

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

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

We know,

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

$⟹$ 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

Question:19 In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is $0.8$ and the probability of passing the second examination is $0.7$. The probability of passing at least one of them is $0.95$. What is the probability of passing both?

Answer:

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

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

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)

$⟹$ 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

Question:20 The probability that a student will pass the final examination in both English and Hindi is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$, what is the probability of passing the Hindi examination?

Answer:

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

$⟹$ P(A $\cup$ B) = 1 - 0.1 = 0.9

Also,

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

$⟹$ P(B) = 0.9 - 0.75 + 0.5 = 0.65

Hence, the probability of passing the Hindi examination is 0.65

Question:21 In a class of $60$ students, $30$ opted for NCC, $32$ opted for NSS and $24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.

(ii) The student has opted neither NCC nor NSS.

(iii) The student has opted NSS but not NCC.

Answer:

Let A be the event that the 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) = $\frac{30}{60}=\frac{1}{2}$

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

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

(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 $\frac{19}{30}$

(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 $\frac{11}{30}$

(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 $\frac{2}{15}$

Question:1 A box contains $10$ red marbles, $20$ blue marbles, and $30$ green marbles. $5$ marbles are drawn from the box, what is the probability that

(i) all will be blue?

(i) atleast one will be green?

Answer:

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 = ${}^{60}{\text{C}}_5$

(i) Number of ways of drawing 5 blue marbles from 20 blue marbles = ${}^{20}{\text{C}}_5$

$\therefore$ Probability of drawing all blue marbles = $\frac{{}^{20}{\text{C}}_5}{{}^{60}{\text{C}}_5}$

(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

⁼ ${}^{(20+10)}{\text{C}}_5{=}^{30}{\text{C}}_5$

$\therefore$ The probability that no marble is green = $\frac{{}^{30}{\text{C}}_5}{{}^{60}{\text{C}}_5}$

$\therefore$ The probability that at least one marble is green = $1-\frac{{}^{30}{\text{C}}_5}{{}^{60}{\text{C}}_5}$

Question:2 $4$ cards are drawn from a well–shuffled deck of $52$ cards. What is the probability of obtaining $3$ diamonds and one spade?

Answer:

Total number of ways of drawing 4 cards from a deck of 52 cards = ${}^{52}{\text{C}}_4$

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

Now, Number of ways of drawing 3 diamonds and 1 spade = ${}^{13}{\text{C}}_3{.}^{13}{\text{C}}_1$

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

$=\frac{{}^{13}{\text{C}}_3{.}^{13}{\text{C}}_1}{{}^{52}{\text{C}}_4}$

Question:3. A die has two faces each with number '1', three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(i) $P(2)$

(ii) P(1 or 3)

(iii) P(not 3)

Answer:

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

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

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