NCERT Solutions for Miscellaneous Exercise Chapter 14 Class 11 - Probability

Question 1: (i) A box contains $\small 10$ red marbles, $\small 20$ blue marbles and $\small 30$ green marbles. $\small 5$ marbles are drawn from the box, what is the probability that all will be blue?

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 = $^{60}\textrm{C}_{5}$

(i) .

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

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

Question 1: (ii) A box contains $\small 10$ red marbles, $\small 20$ blue marbles and $\small 30$ green marbles. $\small 5$ marbles are drawn from the box, what is the probability that 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 = $^{60}\textrm{C}_{5}$

(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 = $1-\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

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

Answer:

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

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

Question 3: (i) 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

$\small P(2)$

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

Question 3: (ii) 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

P(1 or 3)

Answer:

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

Question 3: (iii) 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

P(not 3)

Answer:

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 $\frac{5}{6}$

Question 4: (a) In a certain lottery $\small 10,000$ tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy one ticket

Answer:

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)

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

Question 4: (b) In a certain lottery $\small 10,000$ tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy two tickets

Answer:

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) = $\frac{^{9990}\textrm{C}_{2}}{^{10000}\textrm{C}_{2}}$

Question 4: (c) In a certain lottery $\small 10,000$ tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy $\small 10$ tickets.

Answer:

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) = $\frac{^{9990}\textrm{C}_{10}}{^{10000}\textrm{C}_{10}}$

Question 5: (a) Out of $\small 100$ students, two sections of $\small 40$ and $\small 60$ are formed. If you and your friend are among the $\small 100$ students, what is the probability that

you both enter the same section?

Answer:

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 = $^{100}\textrm{C}_{40}$ (The remaining 60 will automatically be in section B!)

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

$\therefore$ 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 = $= ^{100}\textrm{C}_{60} = ^{100}\textrm{C}_{40}$ (The remaining 40 will automatically be in section A!)

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

$\therefore$ 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 $\frac{17}{33}$

Question 5: (b) Out of $\small 100$ students, two sections of $\small 40$ and $\small 60$ are formed. If you and your friend are among the $\small 100$ students, what is the probability that

you both enter the different sections?

Answer:

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 $\frac{17}{33}$

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

Question 6: Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Answer:

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 = $\frac{4}{3!} = \frac{4}{6} = \frac{2}{3}$

Question 7: (i) A and B are two events such that $\small P(A)=0.54$ , $\small P(B)=0.69$ and $\small P(A\cap B)=0.35$ Find $\small P(A\cup B)$

Answer:

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

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

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

$\implies$ P(A $\cup$ B) = 0.88

Question 7: (ii) A and B are two events such that $\small P(A)=0.54$, $\small P(B)=0.69$ and $\small P({A}\cap {B})=0.35$. Find $\small P({A}'\cap {B}')$

Answer:

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

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

Question 7: (iii) A and B are two events such that $\small P(A)=0.54$ , $\small P(B)=0.69$ and $\small P(A\cap B)=0.35.$ Find $\small P(A\cap B{}')$

Answer:

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

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

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

= 0.54 - 0.35 = 0.19

$\therefore$ P(A $\cap$ B') = 0.19

Question 7: (iv) A and B are two events such that $\small P(A)=0.54$, $\small P(B)=0.69$ and $\small P(A\cap B)=0.35$ . Find $\small P(B\cap A{}')$

Answer:

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

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

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

= 0.69 - 0.35 = 0.34

$\therefore$ P(B $\cap$ A') = 0.34

Question 8: From the employees of a company, $\small 5$ persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

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?

Answer:

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 $\frac{4}{5}$.

Question 9: (i) If $\small 4$-digit numbers greater than $\small 5000$ are randomly formed from the digits $\small 0,1,3.5$ and $\small 7$, what is the probability of forming a number divisible by $\small 5$ when, the digits are repeated?

Answer:

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 =

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

Question 9: (ii) If $\small 4$-digit numbers greater than $\small 5,000$ are randomly formed from the digits $\small 0,1,3,5$ and $\small 7$, what is the probability of forming a number divisible by $\small 5$ when, the repetition of digits is not allowed?

Answer:

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 $^{3}{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 5 and divisible by 5 = $1\times ^{3}{C}_{2}\times2! = 1.3.2 = 6$

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 $^{3}{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 7 and divisible by 5 = $1\times2\times( ^{3}{C}_{2}\times2! )= 1.2.3.2 = 12$

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

Therefore, the required probability =

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

Question 10: The number lock of a suitcase has $\small 4$ wheels, each labelled with ten digits i.e., from $\small 0$ to $\small 9$. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Answer:

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 = $\frac{1}{5040}$