NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 14 Probability - 2025-26

Miscellaneous Exercise

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?

Ans. Total number of marbles \[ = 60\]

The number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .

All the marbles drawn will be blue if we draw \[5\] marbles out of the \[20\] blue marbles. \[5\] blue marbles can be drawn out of the \[20\] blue marbles in \[^{20}{C_5}\] ways.

Probability that all marbles will be blue is \[\frac{{^{20}{C_5}}}{{^{60}{C_5}}}\] .

(ii) at least one will be green?

Ans. Total number of marbles \[ = 60\]

Number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .

The number of ways in which the drawn marbles is not green, that is, when red or blue marbles are drawn, is \[^{20 + 10}{C_5}{ = ^{30}}{C_5}\].

So, probability that no marble is green is \[\frac{{^{30}{C_5}}}{{^{60}{C_5}}}\] .

Therefore, the probability that at least one marble will be green is \[1 - \frac{{^{30}{C_5}}}{{^{60}{C_5}}}\].

2. \[4\] cards are drawn from a well-shuffled deck of \[52\] cards. What is the probability of obtaining three diamonds and one spade?

Ans. Number of ways of drawing \[4\] cards from a pack of \[52\] cards are \[^{52}{C_4}\] .

In a deck, there are \[13\] diamonds and \[13\] spades.

So, the number of ways of drawing three diamonds and one spade is \[^{13}{C_3}{ \times ^{13}}{C_1}\].

Hence, the probability of drawing three diamonds and one spade is \[\frac{{^{13}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_4}}}\].

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

(i) \[P(2)\]

Ans. Total number of faces of the die is \[6\].

The number of faces with the number ‘ \[2\] ’ is \[3\].

Therefore, probability of obtaining the number \[2\] is \[P(2) = \frac{{n(2)}}{{n(S)}}\] .

\[ \Rightarrow P(2) = \frac{3}{6}\]

\[ \Rightarrow P(2) = \frac{1}{2}\]

(ii) P(1 or 3)

Ans. Total number of faces of the die is \[6\].

Number of faces with the number ‘ \[1\] ’ is \[2\] and with the number ‘ \[3\] ’ is \[1\].

Therefore, probability of obtaining the number \[2\] or \[3\] is \[P(1 \cup 3) = \frac{{n(1 \cup 3)}}{{n(S)}}\] .

\[ \Rightarrow P(1 \cup 3) = \frac{{1 + 2}}{6}\] 

\[ \Rightarrow P(1 \cup 3) = \frac{1}{2}\]

(iii) P(not 3)

Ans. Total number of faces of the die is \[6\].

Number of faces with the number ‘ \[3\] ’ is \[1\].

Therefore, probability of obtaining the number \[3\] is \[P(3) = \frac{1}{6}\] .

\[ \Rightarrow P(3') = 1 - P(3)\] 

\[ \Rightarrow P(3') = 1 - \frac{1}{6}\]

\[ \Rightarrow P(3') = \frac{5}{6}\]

4. In a certain lottery, \[10,000\] tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy:

(i) One ticket?

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

If A is the event we buy \[1\] ticket, the probability of winning will be \[P(A) = \frac{{n(A)}}{{n(S)}}\] .

\[ \Rightarrow P(A) = \frac{{10}}{{10,000}}\]

\[ \Rightarrow P(A) = \frac{1}{{1,000}}\] 

So, the probability of not getting a prize is \[P(A') = 1 - P(A)\].

\[ \Rightarrow P(A') = 1 - \frac{1}{{1,000}}\]

\[ \Rightarrow P(A') = \frac{{999}}{{1,000}}\]

(ii) two tickets?

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

Number of tickets on which a prize is not awarded is \[10,000 - 10 = 9,990\].

If B is the event we buy \[2\] tickets, the probability of not winning will be \[P(B) = \frac{{^{9,990}{C_2}}}{{^{10,000}{C_2}}}\] .

(iii) ten tickets?

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

Number of tickets on which a prize is not awarded is \[10,000 - 10 = 9,990\].

If C is the event we buy \[10\] tickets, the probability of not winning will be \[P(C) = \frac{{^{9,990}{C_{10}}}}{{^{10,000}{C_{10}}}}\] .

5. Out of \[100\] students, two sections of \[40\] and \[60\] are formed. If you and your friend are among the \[100\] students, what is the probability that

(i) You both enter the same section?

Ans. You and your friend are among the \[100\] students. 

Total number of ways of selecting \[2\] students out of \[100\] students is ¹⁰⁰C₂.

If both of you enter the same section, you are either among the section having \[40\] students or the one having \[60\] students.

The number of ways in which you both can be in the same section is \[^{40}{C_2}{ + ^{60}}{C_2}\]. 

The probability that you both can be in the same section is \[\frac{{^{40}{C_2}{ + ^{60}}{C_2}}}{{^{100}{C_2}}}\]

\[ \Rightarrow \frac{{\frac{{40!}}{{2!38!}} + \frac{{60!}}{{2!58!}}}}{{\frac{{100!}}{{2!98!}}}}\]

\[ \Rightarrow \frac{{(40 \times 39) + (60 \times 59)}}{{(100 \times 99)}}\]

\[ \Rightarrow \frac{{17}}{{33}}\]

(ii) You both enter different sections?

Ans. The probability that you both can be in the same section is \[\frac{{17}}{{33}}\].

The probability that you both are not in the same section will be \[1 - \frac{{17}}{{33}}\].

\[ \Rightarrow \frac{{16}}{{33}}\]

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.

Ans. Let \[{L_1},{L_2},{L_3}\] be the three letters and \[{E_1},{E_2},{E_3}\] be their corresponding envelopes.

There are \[6\] ways of inserting \[3\] letters in \[3\] envelopes. These are as follows:

\[\begin{array}{l}{L_1}{E_1},{L_2}{E_3},{L_3}{E_2}\\{L_2}{E_2},{L_1}{E_3},{L_3}{E_1}\\{L_3}{E_3},{L_1}{E_2},{L_2}{E_1}\\{L_1}{E_1},{L_2}{E_2},{L_3}{E_3}\\\;{L_1}{E_2},{L_2}{E_3},{L_3}{E_1}\\{L_1}{E_3},{L_2}{E_1},{L_3}{E_2}\end{array}\]

We observe that there are \[4\] ways in which at least one letter is inserted in its proper envelope.

Hence, the required probability is \[\frac{4}{6} = \frac{1}{2}\] .

7. A and B are two events such that \[P(A) = 0.54\] , \[P\left( B \right) = 0.69\] and \[P\left( {A \cap B} \right) = 0.35\] . Find:

(i) \[P\left( {A \cup B} \right)\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

\[ \Rightarrow P(A \cup B) = 0.54 + 0.69 - 0.35\]

\[ \Rightarrow P(A \cup B) = 0.88\]

(ii) \[P\left( {A' \cap B'} \right)\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {A' \cap B'} \right) = P(A \cup B)'\]

\[ \Rightarrow P\left( {A' \cap B'} \right) = 1 - P(A \cup B)\]

\[ \Rightarrow P\left( {A' \cap B'} \right) = 1 - 0.88\]

\[ \Rightarrow P\left( {A' \cap B'} \right) = 0.12\]

(iii) \[P\left( {A \cap B'} \right)\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {A \cap B'} \right) = P\left( A \right)--P\left( {A \cap B} \right)\]

\[ \Rightarrow P\left( {A \cap B'} \right) = 0.54--0.35\]

\[ \Rightarrow P\left( {A \cap B'} \right) = 0.19\]

(iv) \[P\left( {B \cap A'} \right)\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {B \cap A'} \right) = P\left( B \right)--P\left( {A \cap B} \right)\]

\[ \Rightarrow P\left( {B \cap A'} \right) = 0.69--0.35\]

\[ \Rightarrow P\left( {B \cap A'} \right) = 0.34\]

8. From the employees of a company, \[5\] 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 \[35\] years?

Ans. Let E be the event that the spokesperson will be a male and F be the event that the spokesperson will be over \[35\] years of age.

Subsequently, \[P(E) = \frac{3}{5}\] and \[P(F) = \frac{2}{5}\] .

Since there is only one male who is above  \[35\] years of age, \[P(E \cap F) = \frac{1}{5}\] .

We know that \[P(E \cup F) = P(E) + P(F) - P(E \cap F)\]

\[ \Rightarrow P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5}\] 

\[ \Rightarrow P(E \cup F) = \frac{4}{5}\] 

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

9. If four digit numbers greater than 5,000  are randomly formed from the digits \[0\] , \[1\], \[3\], \[5\], and \[7\], what is the probability of forming a number divisible by \[5\] when:

(i) the digits are repeated?

Ans. Since four-digit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . The remaining three digits can be filled by any of the given digits with repetition. 

So, the total number of ways of forming four-digit numbers are \[2 \times 5 \times 5 \times 5\], and subtract \[1\] from the obtained answer, as we cannot count \[5,000\] in the numbers.

\[ \Rightarrow 250 - 1\]

\[ \Rightarrow 249\] ways

A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\] .

Total number of four-digit   numbers   greater than \[5,000\]   that   are   divisible by \[5\] are \[(2 \times 5 \times 5 \times 2) - 1\] .

\[ \Rightarrow 100 - 1\]

\[ \Rightarrow 99\]

Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{99}}{{249}} = \frac{{33}}{{83}}\]

(ii) the repetition of digits is not allowed?

Ans. Since four-digit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . The remaining three digits can be filled by any of the given digits without repetition.

So, the total number of ways of forming four-digit numbers is \[2 \times 4 \times 3 \times 2 = 48\].

A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\].

When the digit at the thousands place is \[5\], the unit place can be filled only with \[0\], so the number of possible ways is \[1 \times 3 \times 2 \times 1 = 6\].

When the digit at the thousands place is \[7\] , the units place can be filled only with \[0\] or \[5\] , so the number of possible ways are \[1 \times 3 \times 2 \times 2 = 12\] .

Total number of four-digit   numbers   greater than \[5,000\]   that   are   divisible by \[5\] are \[12 + 6 = 18\] .

Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{18}}{{48}} = \frac{3}{8}\] .

10. The number lock of a suitcase has \[4\] wheels, each labelled with ten digits i.e., from \[0\] to \[9\] . 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?

Ans. The number lock of a suitcase has \[4\] wheels, each of which is labelled with ten digits, that is, from \[0\] to \[9\].

Number of ways of selecting \[4\] different digits out of \[10\] digits is \[^{10}{C_4}\] .

Further, each combination of \[4\] different digits can be arranged in \[4\] ways.

Number of four digits with no repetitions \[{ = ^{10}}{C_4} \times 4!\]

\[ \Rightarrow \frac{{10!}}{{4!6!}} \times 4!\]

\[ \Rightarrow 10 \times 9 \times 8 \times 7\]

\[ \Rightarrow 5040\]

There is only one number that can open the suitcase.

Thus, the required probability is \[\frac{1}{{5040}}\] .

Conclusion

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 14 on Probability provides a thorough understanding of various probability concepts. Students may develop their understanding and develop their problem-solving abilities by practising these solutions. Because it covers so many various topics and problem-solving categories, this exercise is important to understanding probability. Students who use Vedantu's solutions can take their exams with confidence, understanding that they have studied and identified the basic concepts and strategies required for success in probability.

Class 11 Maths Chapter 14: Exercises Breakdown

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