Anand Classes offers detailed NCERT Solutions for Class 11 Maths Chapter 14 Probability Miscellaneous Exercise to help students master complex probability concepts through clear, step-by-step explanations. These solutions are prepared according to the latest CBSE and NCERT guidelines, making them perfect for exam preparation and quick revision. Each question is solved with proper reasoning, formulas, and examples to enhance conceptual understanding and problem-solving skills. Click the print button to download study material and notes.
NCERT 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? (ii) at least one will be green?
Solution :
Total number of marbles in the box = $10 + 20 + 30 = 60$
Number of ways to draw 5 marbles = ${}^{60}C_{5}$
(i) All will be blue
All marbles will be blue when 5 are drawn from 20 blue marbles.
Number of favourable ways = ${}^{20}C_{5}$
Hence,
$$
P(\text{all blue}) = \frac{{}^{20}C_{5}}{{}^{60}C_{5}}
$$
(ii) At least one will be green
Number of ways to draw marbles that are not green = draw 5 from (red + blue) marbles = $10 + 20 = 30$
So ${}^{30}C_{5}$ ways.
$$
P(\text{no green}) = \frac{{}^{30}C_{5}}{{}^{60}C_{5}}
$$
Therefore,
$$
P(\text{at least one green}) = 1 – \frac{{}^{30}C_{5}}{{}^{60}C_{5}}
$$
✅ Final Answers:
$$
\boxed{
P(\text{all blue}) = \frac{{}^{20}C_{5}}{{}^{60}C_{5}}, \quad
P(\text{at least one green}) = 1 – \frac{{}^{30}C_{5}}{{}^{60}C_{5}}
}
$$
NCERT 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?
Solution :
Total ways to draw 4 cards = ${}^{52}C_{4}$
Number of ways to choose 3 diamonds = ${}^{13}C_{3}$
Number of ways to choose 1 spade = ${}^{13}C_{1}$
Total favourable ways = ${}^{13}C_{3} \times {}^{13}C_{1}$
Hence,
$$
P(\text{3 diamonds and 1 spade}) = \frac{{}^{13}C_{3} \times {}^{13}C_{1}}{{}^{52}C_{4}}
$$
✅ Final Answer:
$$
\boxed{
P = \frac{{}^{13}C_{3} \times {}^{13}C_{1}}{{}^{52}C_{4}}
}
$$
NCERT Question 3 : A die has two faces each with the number ‘1’, three faces each with the number ‘2’, and one face with the number ‘3’. If the die is rolled once, determine
(i) $P(2)$
(ii) $P(1\ \text{or}\ 3)$
(iii) $P(\text{not }3)$
Solution :
Total number of faces = 6
(i) The number of faces with ‘2’ = 3
$$
P(2) = \frac{3}{6} = \frac{1}{2}
$$
(ii) Faces with ‘1’ = 2, with ‘3’ = 1
$$
P(1 \text{ or } 3) = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}
$$
Alternatively,
$$
P(1 \text{ or } 3) = 1 – P(2) = 1 – \frac{1}{2} = \frac{1}{2}
$$
(iii) Faces with ‘3’ = 1
$$
P(3) = \frac{1}{6}, \quad P(\text{not }3) = 1 – \frac{1}{6} = \frac{5}{6}
$$
✅ Final Answers:
$$
\boxed{
P(2) = \frac{1}{2}, \quad P(1 \text{ or } 3) = \frac{1}{2}, \quad P(\text{not }3) = \frac{5}{6}
}
$$
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Explore related topics: probability using combination-based probability, and advanced questions on drawing without replacement.
NCERT Question 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
(a) one ticket
(b) two tickets
(c) ten tickets
Solution :
Total number of tickets = $10000$
Number of prize tickets = $10$
Number of non-prize tickets = $10000-10 = 9990$.
(a) One ticket
Probability of getting a prize with one ticket = $\dfrac{10}{10000}=\dfrac{1}{1000}$.
So probability of not getting a prize with one ticket is
$$
P(\text{no prize with 1 ticket}) = 1 – \frac{1}{1000} = \frac{999}{1000}.
$$
(b) Two tickets
Total ways to choose any 2 tickets : ${}^{10000}C_{2}$.
Ways to choose 2 non-prize tickets : ${}^{9990}C_{2}$.
Hence
$$
P(\text{no prize with 2 tickets})=\frac{{}^{9990}C_{2}}{{}^{10000}C_{2}}.
$$
(c) Ten tickets
Total ways to choose any 10 tickets : ${}^{10000}C_{10}$.
Ways to choose 10 non-prize tickets : ${}^{9990}C_{10}$.
Hence
$$
P(\text{no prize with 10 tickets})=\frac{{}^{9990}C_{10}}{{}^{10000}C_{10}}.
$$
NCERT Question 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
(a) you both enter the same section?
(b) you both enter different sections?
Solution :
Total ways to choose the pair (you and friend) among 100 students: ${}^{100}C_{2}$.
(a) Both enter the same section
To be in the same section, we can either be in 40 students section or in 60 students section.
Ways both are in the 40-student section : ${}^{40}C_{2}$.
Ways both are in the 60-student section : ${}^{60}C_{2}$.
The total number of ways in which we both of us are in same section = 40C2 + 60C2
Hence, the probability of both of us are in same section :
$$
P(\text{same section})=\frac{{}^{40}C_{2}+{}^{60}C_{2}}{{}^{100}C_{2}}.
$$
If you simplify,
$$
\frac{{}^{40}C_{2}+{}^{60}C_{2}}{{}^{100}C_{2}}=\frac{40\cdot39/2+60\cdot59/2}{100\cdot99/2}=\frac{2550}{4950}=\frac{17}{33}.
$$
(b) Both enter different sections
probability of entering the different sections = 1 – P(getting in the same section)
$$
P(\text{different sections})=1-P(\text{same section})=1-\frac{17}{33}=\frac{16}{33}.
$$
NCERT 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.
Solution :
Let the letters be denoted by $L_{1},L_{2},L_{3}$
and the envelopes be denoted by $E_{1},E_{2},E_{3}$.
The letters are inserted into the envelopes and each envelope contains exactly one letter. The combinations that can be made are:
E1L1 E2L2 E3L3
E1L2 E2L1 E3L3
E1L2 E2L3 E3L1
E1L1 E2L3 E3L2
E1L3 E2L1 E3L2
E1L3 E2L2 E3L1
Total permutations (one letter per envelope) = $3! = 6$.
Derangements (no letter in its correct envelope) for $3$ objects is $2$, so number of permutations with at least one correct placement = $6-2=4$.
That is the combinations in which at least one letter is in its proper envelope: E1L1 E2L2 E3L3, E1L3 E2L2 E3L1, E1L1 E2L3 E3L2, E1L2 E2L1 E3L3.
The number of such combinations is 4.
Therefore
$$
P(\text{at least one correct})=\frac{4}{6}=\frac{2}{3}.
$$
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
Related practice: combinatorial probability, lottery probability problems, section-assignment probability, derangements and fixed points.
NCERT Question 7 : A and B are two events such that $P(A) = 0.54$, $P(B) = 0.69$ and $P(A \cap B) = 0.35$. Find:
(i) $P(A \cup B)$
(ii) $P(A’ \cap B’)$
(iii) $P(A \cap B’)$
(iv) $P(B \cap A’)$
Solution:
It is given that:
$$P(A) = 0.54, \quad P(B) = 0.69, \quad P(A \cap B) = 0.35$$
(i) Finding $P(A \cup B)$
We know that,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
Substituting the values,
$$P(A \cup B) = 0.54 + 0.69 – 0.35$$
$$P(A \cup B) = 0.88$$
(ii) Finding $P(A’ \cap B’)$
By De Morgan’s Law,
$$(A’ \cap B’) = (A \cup B)’$$
Hence,
$$P(A’ \cap B’) = 1 – P(A \cup B)$$
Substituting,
$$P(A’ \cap B’) = 1 – 0.88$$
$$P(A’ \cap B’) = 0.12$$
(iii) Finding $P(A \cap B’)$
We know that,
$$P(A \cap B’) = P(A) – P(A \cap B)$$
Substituting,
$$P(A \cap B’) = 0.54 – 0.35$$
$$P(A \cap B’) = 0.19$$
(iv) Finding $P(B \cap A’)$
We know that,
$$P(B \cap A’) = P(B) – P(A \cap B)$$
Substituting,
$$P(B \cap A’) = 0.69 – 0.35$$
$$P(B \cap A’) = 0.34$$
✅ Final Answers:
- (i) $P(A \cup B) = 0.88$
- (ii) $P(A’ \cap B’) = 0.12$
- (iii) $P(A \cap B’) = 0.19$
- (iv) $P(B \cap A’) = 0.34$
For complete NCERT Class 12 Maths Probability Solutions, visit Anand Classes — your trusted institute for JEE, NDA, and CUET preparation.
NCERT Question 8 : From the employees of a company, 5 persons are selected to represent them in the managing committee. The particulars are:
| S.No | Name | Sex | Age (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. Find the probability that the spokesperson will be either male or over 35 years.
Solution :
Total number of employees = $5$
Number of males = $3$ (Harish, Rohan, Salim)
Number of persons over 35 years = $2$ (Sheetal, Salim)
Number of persons who are male and over 35 = $1$ (Salim)
Let $E$ = event “person is male”, $F$ = event “person is over 35”.
Then:
$$
P(E) = \frac{3}{5}, \quad P(F) = \frac{2}{5}, \quad P(E \cap F) = \frac{1}{5}
$$
By the addition rule of probability:
$$
P(E \cup F) = P(E) + P(F) – P(E \cap F)
$$
Substitute the values:
$$
P(E \cup F) = \frac{3}{5} + \frac{2}{5} – \frac{1}{5} = \frac{4}{5}
$$
✅ Final Answer:
$$
\boxed{P(\text{male or over 35}) = \frac{4}{5}}
$$
Download detailed NCERT solutions by Anand Classes for Class 11 & 12 Maths, JEE Main, NDA, and CUET preparation.
Related topics: probability of combined events, addition rule, and age/sex-based probability problems.
NCERT Question 9 : If 4-digit numbers greater than 5,000 are randomly formed from the digits $0, 1, 3, 5, 7$, find the probability of forming a number divisible by 5 when:
(i) digits can be repeated,
(ii) digits cannot be repeated.
Solution :
(i) Digits are repeated
- To form a 4-digit number greater than 5,000, the thousands place can be $5$ or $7$ → 2 choices.
- The remaining 3 digits can be any of the 5 digits (repetition allowed) → $5 \times 5 \times 5 = 125$ ways.
- Total numbers greater than 5,000 = $2 \times 125 = 250$.
Condition for divisibility by 5: A number is divisible by 5 if its one’s place is either filled by 0 or 5. the numbers that can be formed which are divisible by 5 are = 2 x 5 x 5 x 2 = 100
Hence, the probability:
$$
P = \frac{100}{250} = \frac{2}{5}
$$
(ii) Digits are not repeated
- To make 4-digit numbers greater than 5,000 the left-most place, i.e, thousands place can be filled by either 5 or 7.
- The remaining 3 places can be filled by the other 4 digits.
So, the total numbers formed greater than 5,000 when repetition is not allowed are = 2 x 4 x 3 x 2 = 48
A number is divisible by 5 if its one’s place is either filled by 5 or 0.
when the thousand’s place is taken by 5 and one’s place can only be filled with 0 to be divisible by 5. and the rest of the places can be filled by the remaining 3 digits.
the numbers that can be formed starting with 5 and ending at 0 are = 1 x 3 x 2x 1 = 6
and when the thousand’s place is taken by 7 and one’s place can be filled with 0 or 5 to be divisible by 5, the rest of the places can be filled by the remaining 2 digits.
the numbers that can be formed starting with 7 and ending at 5 or 0 are = 1 x 3 x 2 x 2 = 12
the total numbers formed greater than 5,000 and divisible by 5 when repetition of digits are not allowed = 12 + 6 = 18
Hence, probability:
$$
P = \frac{18}{48} = \frac{3}{8}
$$
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
Related topics: probability with digit restrictions, numbers divisible by 5, repetition vs no repetition in combinatorics.
NCERT Question 10 : The number lock of a suitcase has 4 wheels, each labeled with ten digits (0 to 9). The lock opens with a sequence of four digits with no repeats. Find the probability of getting the right sequence.
Solution :
Here, the number lock has 4 wheels each labeled with ten digits i.e from 0 to 9.
- Total digits available: 10 (0 to 9)
- Number of wheels: 4
- No repetition allowed
- Number of ways to select 4 digits out of 10:
$$
^{10}C_4
$$ - Each selected combination of 4 digits can be arranged in 4! ways (since order matters for the lock).
- Total number of possible sequences:
$$
^{10}C_4 \times 4! = 5040
$$ - Only one sequence will open the suitcase.
Hence, the probability of choosing the correct sequence:
$$
P = \frac{1}{5040}
$$
✅ Final Answer:
$$
\boxed{\frac{1}{5040}}
$$
Download detailed NCERT solutions by Anand Classes for Class 11 & 12 Maths, JEE, NDA, and CUET preparation.
Related topics: probability with permutations, sequence selection, and non-repetitive arrangements.
