Anand Classes provides comprehensive NCERT Solutions for Class 11 Maths Chapter 14 Probability Exercise 14.2 to help students strengthen their understanding of theoretical and experimental probability concepts. These step-by-step solutions are designed according to the latest CBSE syllabus and NCERT guidelines, ensuring clarity and accuracy in every answer. Students can use these solutions to prepare effectively for exams and revise important probability formulas and examples with ease. Click the print button to download study material and notes.
NCERT Question 10 : A letter is chosen at random from the word “ASSASSINATION’. Find the probability that the letter is
(i) a vowel (ii) a consonant.
Solution :
The given word ASSASSINATION has a total of 13 letters.
Step 1: Count of vowels and consonants
- Vowels: A, A, A, I, I, O → Total = 6
- Consonants: S, S, S, S, N, T, N → Total = 7
Hence,
$$n(S) = 13$$
(i) Probability of a vowel
Number of favorable outcomes,
$$n(A) = 6$$
Therefore,
$$P(A) = \frac{n(A)}{n(S)} = \frac{6}{13}$$
(ii) Probability of a consonant
Number of favorable outcomes,
$$n(A) = 7$$
Therefore,
$$P(A) = \frac{n(A)}{n(S)} = \frac{7}{13}$$
✅ Final Answers
(i) Probability of vowel = $\boxed{\frac{6}{13}}$
(ii) Probability of consonant = $\boxed{\frac{7}{13}}$
Probability of selecting vowels and consonants from a word, Class 11 Maths NCERT Chapter 14, letter selection probability examples, Anand Classes NCERT solutions for JEE Main, NDA, and CUET preparation.
NCERT 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.]
Solution :
Total numbers available in the draw = 20
Numbers to be selected = 6
Therefore, number of ways to select 6 numbers from 20 numbers :-
$$n(S) = {^{20}C_6}$$
Let the event A be “the chosen six numbers match with the six numbers already fixed by the lottery committee.”
Number of favorable outcomes,
$$n(A) = {^{6}C_6} = 1$$
Hence, the probability of winning the prize is
$$
P(A) = \frac{n(A)}{n(S)} = \frac{{^{6}C_6}}{{^{20}C_6}}
$$
Substituting the values,
$$
P(A) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 14!}{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14!}
$$
Simplifying,
$$
P(A) = \frac{1}{38760}
$$
✅ Final Answer
The probability of winning the prize is
$$\boxed{\frac{1}{38760}}$$
Probability of winning a lottery problem, combinations and probability Class 11 Maths NCERT Chapter 14, Anand Classes NCERT Solutions for JEE Main, NDA, and CUET preparation.
NCERT 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) $P(A) = 0.5,\ P(B) = 0.4,\ P(A \cup B) = 0.8$
Solution :
(i)
Given:
$$P(A) = 0.5, \quad P(B) = 0.7, \quad P(A \cap B) = 0.6$$
Since for any two events,
$$P(A \cap B) \leq P(A) \text{ and } P(A \cap B) \leq P(B)$$
But here,
$$P(A \cap B) = 0.6 > P(A) = 0.5$$
Therefore, the given probabilities are not consistently defined.
(ii)
Given:
$$P(A) = 0.5, \quad P(B) = 0.4, \quad P(A \cup B) = 0.8$$
Using the formula,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
Substituting values,
$$0.8 = 0.5 + 0.4 – P(A \cap B)$$
$$P(A \cap B) = 0.9 – 0.8 = 0.1$$
Since $P(A \cap B) < P(A)$ and $P(A \cap B) < P(B)$,
the given probabilities are consistently defined.
NCERT Question 13 : Fill in the blanks in the following table:
| Case | P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
|---|---|---|---|---|
| (i) | 1/3 | 1/5 | 1/15 | ……… |
| (ii) | 0.35 | …. | 0.25 | 0.6 |
| (iii) | 0.5 | 0.35 | ……… | 0.7 |
Solution :
(i)
Given:
$$P(A) = \frac{1}{3}, \quad P(B) = \frac{1}{5}, \quad P(A \cap B) = \frac{1}{15}$$
Using,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$P(A \cup B) = \frac{1}{3} + \frac{1}{5} – \frac{1}{15}$$
$$P(A \cup B) = \frac{5 + 3 – 1}{15} = \frac{7}{15}$$
Hence,
$$P(A \cup B) = \frac{7}{15}$$
(ii)
Given:
$$P(A) = 0.35, \quad P(A \cap B) = 0.25, \quad P(A \cup B) = 0.6$$
Using,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$0.6 = 0.35 + P(B) – 0.25$$
$$P(B) = 0.6 + 0.25 – 0.35 = 0.5$$
Hence,
$$P(B) = 0.5$$
(iii)
Given:
$$P(A) = 0.5, \quad P(B) = 0.35, \quad P(A \cup B) = 0.7$$
Using,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$0.7 = 0.5 + 0.35 – P(A \cap B)$$
$$P(A \cap B) = 0.85 – 0.7 = 0.15$$
Hence,
$$P(A \cap B) = 0.15$$
✅ Final Answers:
| Case | P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
|---|---|---|---|---|
| (i) | 1/3 | 1/5 | 1/15 | 7/15 |
| (ii) | 0.35 | 0.5 | 0.25 | 0.6 |
| (iii) | 0.5 | 0.35 | 0.15 | 0.7 |
Probability formula for union and intersection of two events, NCERT Class 11 Chapter 16 solutions by Anand Classes, JEE and CUET probability concepts, consistent probability conditions explained.
NCERT Question 14 : Given $P(A) = \dfrac{3}{5}$ and $P(B) = \dfrac{1}{5}$. Find $P(A \text{ or } B)$, if $A$ and $B$ are mutually exclusive events.
Solution :
For mutually exclusive events,
$$P(A \cup B) = P(A) + P(B)$$
Substitute the values:
$$P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$$
✅ Final Answer
$$\boxed{P(A \text{ or } B) = \frac{4}{5}}$$
NCERT Question 15 : If $E$ and $F$ are events such that $P(E) = \dfrac{1}{4}$, $P(F) = \dfrac{1}{2}$ and $P(E \text{ and } F) = \dfrac{1}{8}$, find
(i) $P(E \text{ or } F)$, (ii) $P(\text{not }E \text{ and not }F)$
Solution :
(i) We know that
$$P(E \cup F) = P(E) + P(F) – P(E \cap F)$$
Substitute the given values:
$$P(E \cup F) = \frac{1}{4} + \frac{1}{2} – \frac{1}{8}$$
Simplify:
$$P(E \cup F) = \frac{2}{8} + \frac{4}{8} – \frac{1}{8} = \frac{5}{8}$$
Hence,
$$P(E \text{ or } F) = \frac{5}{8}$$
(ii)
Now,
$$P(E’ \cap F’) = P((E \cup F)’) = 1 – P(E \cup F)$$
Substitute the value of $P(E \cup F)$:
$$P(E’ \cap F’) = 1 – \frac{5}{8} = \frac{3}{8}$$
✅ Final Answers
(i) $P(E \text{ or } F) = \frac{5}{8}$
(ii) $P(\text{not }E \text{ and not }F) = \frac{3}{8}$
NCERT Question 16 : Events $E$ and $F$ are such that $P(\text{not }E \text{ or not }F) = 0.25$. State whether $E$ and $F$ are mutually exclusive.
Solution :
Given:
$$P(E’ \cup F’) = 0.25$$
We know that
$$P(E’ \cup F’) = P((E \cap F)’)$$
Thus,
$$1 – P(E \cap F) = 0.25$$
$$P(E \cap F) = 0.75$$
For events to be mutually exclusive, $P(E \cap F)$ must be $0$.
But here, $P(E \cap F) = 0.75 \neq 0$.
✅ Final Answer
Hence, $E$ and $F$ are not mutually exclusive events.
Mutually exclusive events probability rule, probability of union and intersection, complement of events, NCERT Class 11 Maths Chapter 16 by Anand Classes for JEE Main, NDA, and CUET preparation.
NCERT Question 17 : A and B are events such that $P(A) = 0.42$, $P(B) = 0.48$ and $P(A \text{ and } B) = 0.16$. Determine
(i) $P(\text{not }A)$, (ii) $P(\text{not }B)$, and (iii) $P(A \text{ or } B)$.
Solution :
(i)
$$P(\text{not }A) = 1 – P(A)$$
$$P(\text{not }A) = 1 – 0.42 = 0.58$$
(ii)
$$P(\text{not }B) = 1 – P(B)$$
$$P(\text{not }B) = 1 – 0.48 = 0.52$$
(iii)
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$P(A \cup B) = 0.42 + 0.48 – 0.16 = 0.74$$
✅ Final Answers
(i) $P(\text{not }A) = 0.58$
(ii) $P(\text{not }B) = 0.52$
(iii) $P(A \text{ or } B) = 0.74$
NCERT Question 18 : In Class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% 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.
Solution :
Let $A$ be the event that the student studies Mathematics,
and $B$ be the event that the student studies Biology.
Then,
$$P(A) = \frac{40}{100} = \frac{2}{5}, \quad P(B) = \frac{30}{100} = \frac{3}{10}, \quad P(A \cap B) = \frac{10}{100} = \frac{1}{10}$$
Now,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$P(A \cup B) = \frac{2}{5} + \frac{3}{10} – \frac{1}{10}$$
$$P(A \cup B) = \frac{4}{10} + \frac{3}{10} – \frac{1}{10} = \frac{6}{10} = \frac{3}{5}$$
✅ Final Answer
$$P(\text{studying Mathematics or Biology}) = \boxed{\frac{3}{5}}$$
NCERT Question 19 : In an entrance test 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?
Solution :
Let $A$ = event of passing the first examination
and $B$ = event of passing the second examination.
Given:
$$P(A) = 0.8, \quad P(B) = 0.7, \quad P(A \cup B) = 0.95$$
Using the formula,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
Substitute the values:
$$0.95 = 0.8 + 0.7 – P(A \cap B)$$
Simplify:
$$P(A \cap B) = 1.5 – 0.95 = 0.55$$
✅ Final Answer
The probability of passing both examinations is
$$\boxed{0.55}$$
Probability of union and intersection of events, complement probability formulas, Class 11 Maths Chapter 16 NCERT Solutions, Anand Classes Probability Problems for JEE Main, NDA, and CUET preparation.
NCERT 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?
Solution:
Given,
$P(A) = 0.75$, $P(A \cap B) = 0.5$, and $P(A’ \cap B’) = 0.1$
We know that,
$$P(A’ \cap B’) = 1 – P(A \cup B)$$
Hence,
$$P(A \cup B) = 1 – P(A’ \cap B’)$$
$$= 1 – 0.1$$
$$= 0.9$$
Now,
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
Substitute the given values:
$$0.9 = 0.75 + P(B) – 0.5$$
Therefore,
$$P(B) = 0.9 + 0.5 – 0.75$$
$$= 0.65$$
Final Answer:
$$\boxed{P(B) = 0.65}$$
NCERT 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.
Solution:
Total number of students in the class = 60
$$n(S) = 60$$
Let NCC be event $A$ and NSS be event $B$.
Given:
$$n(A) = 30, \quad n(B) = 32, \quad n(A \cap B) = 24$$
So,
$$P(A) = \frac{n(A)}{n(S)} = \frac{30}{60} = \frac{1}{2}$$
$$P(B) = \frac{n(B)}{n(S)} = \frac{32}{60} = \frac{8}{15}$$
$$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{24}{60} = \frac{2}{5}$$
(i) The student opted for NCC or NSS
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
$$P(A \cup B) = \frac{1}{2} + \frac{8}{15} – \frac{2}{5}$$
$$P(A \cup B) = \frac{15 + 16 – 12}{30}$$
$$P(A \cup B) = \frac{19}{30}$$
Hence,
$$\boxed{P(A \cup B) = \frac{19}{30}}$$
(ii) The student opted neither NCC nor NSS
$$P(A’ \cap B’) = 1 – P(A \cup B)$$
$$P(A’ \cap B’) = 1 – \frac{19}{30}$$
$$P(A’ \cap B’) = \frac{11}{30}$$
Hence,
$$\boxed{P(A’ \cap B’) = \frac{11}{30}}$$
(iii) The student opted NSS but not NCC
$$n(B – A) = n(B) – n(A \cap B)$$
$$n(B – A) = 32 – 24 = 8$$
So,
$$P(B – A) = \frac{n(B – A)}{n(S)} = \frac{8}{60} = \frac{2}{15}$$
Hence,
$$\boxed{P(B – A) = \frac{2}{15}}$$
Final Answers:
(i) $\frac{19}{30}$
(ii) $\frac{11}{30}$
(iii) $\frac{2}{15}$
Explore more conceptual probability examples and download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
