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CBSE Important Questions Class 12 Maths Chapter 13 – Probability

Class 12 Maths Chapter 13 Probability is part of the CBSE Class 12 Maths syllabus. It goes beyond the basic probability of earlier classes. Students learn conditional probability, the multiplication theorem, independent events, the total probability theorem, Bayes' theorem, random variables and their probability distributions, and the mean of a random variable.

These CBSE important questions for Class 12 Maths Chapter 13 cover conditional probability, independent events, the total probability theorem, Bayes' theorem applied to a problem of identifying the source, probability distributions, and finding the mean and variance of a random variable. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define conditional probability and write its formula.

Answer: Conditional probability is the probability of an event A happening given that another event B has already happened. It is written P(A|B).
The formula is
P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0
The idea is that once B has happened, the sample space shrinks to B, so we measure A only within B.

Q2. If P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.2, find P(A|B) and P(B|A).

Answer: P(A|B) = P(A ∩ B) / P(B)
= 0.2 / 0.3
= 2/3
P(B|A) = P(A ∩ B) / P(A)
= 0.2 / 0.6
= 1/3
Note that P(A|B) and P(B|A) are different — a common source of mistakes.

Q3. State the multiplication theorem of probability. When are two events called independent?

Answer: Multiplication theorem:
P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
Independent events: Two events A and B are independent if the happening of one does not affect the probability of the other, that is P(A|B) = P(A).
In that case the multiplication theorem simplifies to
P(A ∩ B) = P(A) × P(B)
This product rule is the test used to check independence.

Q4. A die is thrown twice. Find the probability that the sum is 6 given that the first throw showed 2.

Answer: Let A be "the sum is 6" and B be "the first throw is 2".
P(B): The first throw being 2 gives 6 outcomes out of 36, so P(B) = 6/36 = 1/6.
P(A ∩ B): For both to happen, the throws must be (2, 4), which is 1 outcome out of 36.
P(A ∩ B) = 1/36
P(A|B) = (1/36) / (6/36)
= 1/6

Q5. State the theorem of total probability.

Answer: Let E₁, E₂, ..., Eₙ be a set of events that are mutually exclusive and exhaustive (they do not overlap and together they cover the whole sample space), each with non-zero probability. Then for any event A:
P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ... + P(Eₙ)P(A|Eₙ)
In words: the probability of A is the sum, over every possible "route" Eᵢ, of the probability of taking that route times the probability of A along it.

Q6. State Bayes' theorem.

Answer: Let E₁, E₂, ..., Eₙ be mutually exclusive and exhaustive events, and let A be an event with P(A) ≠ 0. Then
P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ)
Bayes' theorem lets us work backwards: we know the outcome A has happened, and we want the probability that it came from a particular source Eᵢ. The P(Eᵢ) are called prior probabilities and P(Eᵢ|A) the posterior probability.

Q7. Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Answer: Let E₁ = ball from Bag I, E₂ = ball from Bag II, and A = the ball is red.
P(E₁) = P(E₂) = 1/2
P(A|E₁) = 3/7 and P(A|E₂) = 5/11
By Bayes' theorem:
P(E₂|A) = [P(E₂)P(A|E₂)] / [P(E₁)P(A|E₁) + P(E₂)P(A|E₂)]
= (1/2 × 5/11) / (1/2 × 3/7 + 1/2 × 5/11)
= (5/22) / (3/14 + 5/22)
Taking the LCM 154: = (35/154) / (33/154 + 35/154)
= 35/68
So the probability is 35/68.

Q8. What is a random variable and a probability distribution?

Answer: A random variable X is a function that assigns a real number to each outcome of a random experiment.
For example, if two coins are tossed and X is the number of heads, then X can take the values 0, 1 or 2.
A probability distribution of X is a table listing each possible value of X with its probability.
Two conditions must always hold:
• each pᵢ ≥ 0
• the sum of all the probabilities is 1

Q9. Two cards are drawn at random without replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer: Let X be the number of aces drawn, so X can be 0, 1 or 2.
P(X = 0) = (48/52) × (47/51) = 2256/2652 = 188/221
P(X = 1) = 2 × (4/52) × (48/51) = 384/2652 = 32/221
P(X = 2) = (4/52) × (3/51) = 12/2652 = 1/221
Check: 188/221 + 32/221 + 1/221 = 221/221 = 1 ✓

Q10. Write the formula for the mean and variance of a random variable. Find the mean number of heads when three coins are tossed.

Answer: Mean (expectation): μ = E(X) = Σ xᵢ pᵢ
Variance: σ² = Σ xᵢ²pᵢ − μ²
For three coins: X can be 0, 1, 2 or 3 heads, with probabilities 1/8, 3/8, 3/8 and 1/8.
Mean = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8)
= 0 + 3/8 + 6/8 + 3/8
= 12/8
= 1.5 heads

Q11. A problem in Maths is given to three students whose chances of solving it are 1/2, 1/3 and 1/4. What is the probability that the problem will be solved?

Answer: It is easier to find the probability that none of them solves it, and then subtract from 1.
P(first fails) = 1 − 1/2 = 1/2
P(second fails) = 1 − 1/3 = 2/3
P(third fails) = 1 − 1/4 = 3/4
Since the three work independently:
P(none solves) = (1/2)(2/3)(3/4) = 6/24 = 1/4
P(problem is solved) = 1 − 1/4 = 3/4

Q12. A factory has two machines A and B. Machine A produces 60% of the items and machine B produces 40%. Of these, 2% from A and 1% from B are defective. An item is picked at random and found to be defective. What is the probability that it came from machine A?

Answer: Let E₁ = item from A, E₂ = item from B, and D = item is defective.
P(E₁) = 0.6 and P(E₂) = 0.4
P(D|E₁) = 0.02 and P(D|E₂) = 0.01
By the total probability theorem:
P(D) = (0.6)(0.02) + (0.4)(0.01) = 0.012 + 0.004 = 0.016
By Bayes' theorem:
P(E₁|D) = 0.012 / 0.016
= 12/16
= 3/4
So there is a 75% chance the defective item came from machine A.

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Q1.

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4 . What is the probability that the student knows the answer given that he answered it correctly

Opt.

Let E₁= Student knows the answer
E₂= Student guesses the answer
And A = answer is correct

Ans.

Let E₁= Student knows the answer
E₂= Student guesses the answer
And A = answer is correct

$\begin{array}{l} {P(E}_{1}) = \frac{3}{4} {P(A | E}_{1}) = 1 \\ {P(E}_{2}) = \frac{1}{4} {P(A | E}_{2}) = \frac{1}{4} \\ Thus, by Bayes’s theorem, \\ {P(E}_{1} | A) = \frac{{P(E}_{1} {)P(A | E}_{1})}{{P(E}_{1} {)P(A | E}_{1}) + {P(E}_{2} {)P(A | E}_{2})} \\ = \frac{\frac{3}{4} — 1}{\frac{3}{4} — 1 + \frac{1}{4} — \frac{1}{4}} = \frac{12}{13} . \end{array}$

Q2.

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Opt.

Let S₁ = six occurs

S₂ = six does not occur.

and E = the man reports that six occurs in the throwing of the die

${P(S}_{1}) = \frac{1}{6} {P(E | S}_{1}) = \frac{3}{4} {P(S}_{2}) = \frac{5}{6} {P(E | S}_{2}) = \frac{1}{4} Using Bayes’s theorem, {P(S}_{1} | E) = {P(S}_{1} {)P(E | S}_{1})$

Ans.

Let S₁ = six occurs

S₂ = six does not occur.

and E = the man reports that six occurs in the throwing of the die

$\begin{array}{l} {P(S}_{1}) = \frac{1}{6} {P(E | S}_{1}) = \frac{3}{4} \\ {P(S}_{2}) = \frac{5}{6} {P(E | S}_{2}) = \frac{1}{4} \\ Using Bayes’s theorem, \\ {P(S}_{1} | E) = \frac{{P(S}_{1} {)P(E | S}_{1})}{{P(S}_{1} {)P(E | S}_{1}) + {P(S}_{2} {)P(E | S}_{2})} \\ = \frac{\frac{1}{6} — \frac{3}{4}}{\frac{1}{6} — \frac{3}{4} + \frac{5}{6} — \frac{1}{4}} = \frac{3}{8} . \end{array}$

Q3.

Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Opt.

Let E₁ = choosing the bag I, E₂ = choosing the bag II and A = drawing a red ball

$\begin{array}{l} \end{array}$ $\begin{array}{l} {P(E}_{1}) = \frac{1}{2} {P(A | E}_{1}) = P (drawing a red ball from bag I) = \frac{3}{7} \\ {P(E}_{2}) = \frac{1}{2} {P(A | E}_{2}) = P (drawing a red ball from bag II) = \frac{5}{11} \\ \begin{array}{l} Using Bayes’s theorem, \end{array} \end{array}$

Ans. Let E₁ = choosing the bag I, E₂ = choosing the bag II and A = drawing a red ball.

$\begin{array}{l} {P(E}_{1}) = \frac{1}{2} {P(A | E}_{1}) = P (drawing a red ball from bag I) = \frac{3}{7} \\ {P(E}_{2}) = \frac{1}{2} {P(A | E}_{2}) = P (drawing a red ball from bag II) = \frac{5}{11} \\ \begin{array}{l} Using Bayes’s theorem, \\ {P(E}_{2} | A) = \frac{{P(E}_{2} {)P(A | E}_{2})}{{P(E}_{1} {)P(A | E}_{1}) + {P(E}_{2} {)P(A | E}_{2})} \end{array} \\ = \frac{\frac{1}{2} — \frac{5}{11}}{\frac{2}{2} — \frac{3}{7} + \frac{1}{2} — \frac{5}{11}} \\ = \frac{35}{68} . \end{array}$

CBSE Important Questions Class 12 Maths Chapter 13 – Probability

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