Probability MCQ with Solutions | Class 12

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Probability MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 9 important questions step by step with clear explanations. You can also watch the video solutions for better understanding. This is in continuation of the previous topic on Vector Algebra MCQs.

Ready? Okay, let’s begin with the first question!

Probability MCQ#1 (NCERT Exercise 13.1 – Question 16)

📍If $P(A)=\dfrac{1}{2}$ , $P(B)=0$ , then $P(A|B)$ is

(A) $0$
(B) $\dfrac{1}{2}$
(C) not defined
(D) $1$

Answer:
✅ Correct option: (C)

Explanation:
Conditional probability is defined as

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

Given $P(B)=0$ .

So the denominator becomes zero and the expression is not defined.

Hence, $P(A|B)$ is not defined.

Therefore, option (C) is correct.

Watch Video Solution on YouTube

Probability MCQ#2 (NCERT Exercise 13.1 – Question 17)

📍If $A$ and $B$ are events such that $P(A|B)=P(B|A)$ , then

(A) $A\subset B \text{ but } A\neq B$
(B) $A=B$
(C) $A\cap B=\phi$
(D) $P(A)=P(B)$

Answer:
✅ Correct option: (D)

Explanation:
$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

Given
$P(A|B)=P(B|A)$

So,
$\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A\cap B)}{P(A)}$

Cancelling $P(A\cap B)$ (assuming it is non-zero), we get

$P(A)=P(B)$

Hence option (D) is correct.

Watch Video Solution on YouTube

Let me tell you all of these 9 Probability MCQ are very simple and easy to understand and therefore scoring too. But you must practice them regularly.

So, my suggestion to you is to solve all of these Probability MCQ from NCERT a good number of times and feel confident. And if you ever need any clarification, just drop a comment — I’ll be more than happy to help.

Probability MCQ#3 (NCERT Exercise 13.2 – Question 17)

📍The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) $0$
(B) $\dfrac{1}{3}$
(C) $\dfrac{1}{12}$
(D) $\dfrac{1}{36}$

Answer:
✅ Correct option: (D)

Explanation:
The only even prime number is

$2$

When two dice are rolled, total possible outcomes are

$6\times6=36$

For each die to show an even prime number, both must show

$2$

Favourable outcome = $(2,2)$

Number of favourable outcomes = $1$

Therefore, $P=\dfrac{1}{36}$

Hence option (D) is correct.

Watch Video Solution on YouTube

Probability MCQ#4 (NCERT Exercise 13.2 – Question 18)

📍Two events $A$ and $B$ will be independent, if

(A) $A \text{ and } B \text{ are mutually exclusive}$
(B) $P(A'B')=[1-P(A)][1-P(B)]$
(C) $P(A)=P(B)$
(D) $P(A)+P(B)=1$

Answer:
✅ Correct option: (B)

Explanation:
Two events are independent if

$P(A\cap B)=P(A)P(B)$

For independent events, complements are also independent.

So, $P(A' \cap B')=P(A')P(B')$

Now, $P(A')=1-P(A),\quad P(B')=1-P(B)$

Therefore, $P(A'B')=[1-P(A)][1-P(B)]$

Hence option (B) is correct.

Video Solution on YouTube

Did you know, there are about 100 MCQs in NCERT textbooks for Class 12 (both Part 1 and Part 2)? I have been taking one Chapter at a time and showing you the explanations for each of the questions. Infact this very topic on Probability MCQ is the last one in this series and so we are completing all the 100 MCQs today, wow!

I have mentioned in the earlier topics as well, that I have covered these solutions in the video form also and they are freely available on my YouTube channel, @Mathsbetter.

And here also, in each of the 9 Probability MCQ on this page, I have provided the direct links to each and every MCQ for you to understand the solutions in a very simple and easy manner.

Probability MCQ#5 (NCERT Exercise 13.3 – Question 13)

📍Probability that $A$ speaks truth is $\dfrac{4}{5}$ . A coin is tossed. $A$ reports that a head appears. The probability that actually there was head is

(A) $\dfrac{4}{5}$
(B) $\dfrac{1}{2}$
(C) $\dfrac{1}{5}$
(D) $\dfrac{2}{5}$

Answer:
✅ Correct option: (A)

Explanation:
Let $H$ denote that the coin shows head and $E$ denote that $A$ reports head.

$P(H)=\dfrac{1}{2}$ and $P(T)=\dfrac{1}{2}$

$A$ speaks truth with probability
$P(\text{truth})=\dfrac{4}{5}$

Therefore,
$P(E|H)=\dfrac{4}{5}$ (he reports correctly)
$P(E|T)=\dfrac{1}{5}$ (he lies)

Using Bayes’ theorem, $P(H|E)=\dfrac{P(H)P(E|H)}{P(H)P(E|H)+P(T)P(E|T)}$

$=\dfrac{\frac{1}{2}\times\frac{4}{5}}{\frac{1}{2}\times\frac{4}{5}+\frac{1}{2}\times\frac{1}{5}}$

$=\dfrac{\frac{4}{10}}{\frac{5}{10}}=\dfrac{4}{5}$

Hence option (A) is correct.

Watch Video Solution on YouTube

Probability MCQ#6 (NCERT Exercise 13.3 – Question 14)

📍If $A$ and $B$ are two events such that $A\subset B$ and $P(B)\ne0$ , then which of the following is correct?

(A) $P(A|B)=\dfrac{P(B)}{P(A)}$
(B) $P(A|B) < P(A)$
(C) $P(A|B) \ge P(A)$
(D) None of these

Answer:
✅ Correct option: (C)

Explanation:
Given
$A\subset B$

Therefore, $A\cap B=A$

Using conditional probability, $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$P(A|B)=\dfrac{P(A)}{P(B)}$

Since $P(B)\le1$ , we have $\dfrac{P(A)}{P(B)}\ge P(A)$

Thus, $P(A|B)\ge P(A)$

Hence option (C) is correct.

Watch Video Solution on Yoube

You may have been following this website as a resource for some of the important questions. For example, how to Integrate Square Root of tan x, or may be a complete guide to look at using Matrix Method in solving a system of linear equations.

Then trust me, similarly, this topic on NCERT Class 12 Probability MCQ is going to help you in many ways.

Probability MCQ#7 (NCERT Misc. Exercise Chapter 13 – Question 11)

📍If $A$ and $B$ are two events such that $P(A)\ne0$ and $P(B|A)=1$ , then

(A) $A\subset B$
(B) $B\subset A$
(C) $B=\phi$
(D) $A=\phi$

Answer:
✅ Correct option: (A)

Explanation:
Given
$P(B|A)=1$

Using conditional probability, $P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

So, $\dfrac{P(A\cap B)}{P(A)}=1$

$P(A\cap B)=P(A)$

This happens only when all outcomes of $A$ are contained in $B$ . Therefore, $A\subset B$

Hence option (A) is correct.

Watch Video Solution on YouTube

Probability MCQ#8 (NCERT Misc. Exercise Chapter 13 – Question 12)

📍If $P(A|B) > P(A)$ , then which of the following is correct :

(A) $P(B|A) < P(B)$
(B) $P(A\cap B) < P(A)P(B)$
(C) $P(B|A) > P(B)$
(D) $P(B|A) = P(B)$

Answer:
✅ Correct option: (C)

Explanation:
Given $P(A|B) > P(A)$

Using conditional probability, $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

So, $\dfrac{P(A\cap B)}{P(B)} > P(A)$

Multiplying both sides by $P(B)$ , we get $P(A\cap B) > P(A)P(B)$

And now dividing both sides by $P(A)$ , we get $\dfrac{P(A\cap B)}{P(A)} > P(B)$

But $\dfrac{P(A\cap B)}{P(A)} = P(B|A)$ , Therefore, $P(B|A) > P(B)$

Hence option (C) is correct.

Watch Video Solution on YouTube

Probability MCQ#9 (NCERT Misc. Exercise Chapter 13 – Question 13)

📍If $A$ and $B$ are any two events such that $P(A)+P(B)-P(A\cap B)=P(A)$ , then

(A) $P(B|A)=1$
(B) $P(A|B)=1$
(C) $P(B|A)=0$
(D) $P(A|B)=0$

Answer:
✅ Correct option: (B)

Explanation:
Given $P(A)+P(B)-P(A\cap B)=P(A)$

Cancelling $P(A)$ from both sides, we get $P(B)-P(A\cap B)=0$

i.e. $P(A\cap B) = P(B)$

Now, dividing both sides by $P(B)$ , we get $\dfrac{P(A\cap B)}{P(B)} =\dfrac{P(B)}{P(B)} = 1$

or $P(A|B)=1$

Thus event $B$ always implies event $A$ . Hence option (B) is correct.

Watch Video Solution on YouTube

Quick Answer Key

Q1 → (C)
Q2 →(D)
Q3 →(D)
Q4 → (B)
Q5 → (A)
Q6 → (C)
Q7 → (A)
Q8 → (C)
Q9 → (B)

Closing Note

That completes, not only the important NCERT Class 12 Maths Probability MCQ from chapter 13, but also, all the 100 MCQs from NCERT textbook for Class 12.
I hope the explanations and video solutions made things clearer and gave you more confidence. Stay tuned for the next topics or series of important exercises and questions from NCERT textbook.

👉 Make sure to practice these questions again to strengthen your concepts.
👉 Watch the video solutions for a clearer and faster revision.

Keep practicing, and you’ll master the probability easily! 🚀

Probability: Chapter 13 Links

Probability MCQ#1 (NCERT Exercise 13.1 – Question 16)

Probability MCQ#2 (NCERT Exercise 13.1 – Question 17)

Probability MCQ#3 (NCERT Exercise 13.2 – Question 17)

Probability MCQ#4 (NCERT Exercise 13.2 – Question 18)

Probability MCQ#5 (NCERT Exercise 13.3 – Question 13)

Probability MCQ#6 (NCERT Exercise 13.3 – Question 14)

Probability MCQ#7 (NCERT Misc. Exercise Chapter 13 – Question 11)

Probability MCQ#8 (NCERT Misc. Exercise Chapter 13 – Question 12)

Probability MCQ#9 (NCERT Misc. Exercise Chapter 13 – Question 13)

Quick Answer Key

Closing Note

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