CUET UG Mathematics Probability Probability Conditional Bayes Theorem Binomial Distribution

Must‑Know (15–20 detailed bullets)

• Conditional probability of event A given B is defined as ( P(A|B) = \frac{P(A \cap B)}{P(B)} ), provided ( P(B) > 0 ); e.g., if two dice are rolled and sum is 7, probability that one die shows 4 is ( \frac{2/36}{6/36} = \frac{1}{3} ).
• If A and B are independent events, then ( P(A \cap B) = P(A) \cdot P(B) ), and ( P(A|B) = P(A) ); e.g., tossing a fair coin twice: ( P(H_1 \cap H_2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} ).
• Total Probability Theorem: If ( E_1, E_2, ..., E_n ) are mutually exclusive and exhaustive events, then for any event A, ( P(A) = \sum_{i=1}^{n} P(E_i)P(A|E_i) ); e.g., 60% boys and 40% girls in a class, 80% of boys and 60% of girls pass — total pass probability = ( 0.6 \times 0.8 + 0.4 \times 0.6 = 0.72 ).
• Bayes’ Theorem: ( P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum_{j=1}^{n} P(E_j)P(A|E_j)} ); used to find reverse probability; e.g., probability a defective item came from machine X when machines X and Y produce 60% and 40% items with defect rates 2% and 3% respectively.
• In Bayes’ Theorem, the events ( E_i ) are called hypotheses, ( P(E_i) ) are prior probabilities, and ( P(E_i|A) ) are posterior probabilities.
• A random variable is a real-valued function defined on the sample space; e.g., number of heads in 3 coin tosses takes values 0, 1, 2, 3.
• Probability distribution of a discrete random variable lists all possible values with corresponding probabilities summing to 1; e.g., for a fair die: ( P(X = x) = \frac{1}{6}, x = 1,2,...,6 ).
• Mean (expected value) of a random variable X: ( \mu = E(X) = \sum x_i p_i ); e.g., for a fair die: ( E(X) = \frac{1+2+3+4+5+6}{6} = 3.5 ).
• Variance of X: ( Var(X) = \sigma^2 = \sum (x_i - \mu)^2 p_i = E(X^2) - [E(X)]^2 ); e.g., variance of fair die: ( E(X^2) = \frac{91}{6}, \mu = 3.5 \Rightarrow Var(X) = \frac{91}{6} - (3.5)^2 = \frac{35}{12} \approx 2.92 ).
• Bernoulli trial has only two outcomes: success (S) with probability p and failure (F) with probability q = 1 – p; e.g., tossing a coin once with P(H) = p.
• Binomial distribution applies when there are n independent Bernoulli trials; probability of exactly x successes: ( P(X = x) = ^nC_x p^x (1-p)^{n-x} ), where ( x = 0,1,2,...,n ).
• For binomial distribution, mean = np, variance = np(1–p); e.g., 10 coin tosses (p = 0.5): mean = 5, variance = 2.5.
• Standard deviation of binomial distribution: ( \sqrt{np(1-p)} ); e.g., n = 100, p = 0.2 → SD = ( \sqrt{100 \times 0.2 \times 0.8} = 4 ).
• If ( P(A) = 0.4, P(B) = 0.8, P(A \cap B) = 0.32 ), then A and B are independent because ( 0.4 \times 0.8 = 0.32 ).
• If ( P(A|B) = P(B|A) ), it does not imply ( P(A) = P(B) ); counterexample: ( P(A) = 0.3, P(B) = 0.6, P(A \cap B) = 0.18 \Rightarrow P(A|B) = 0.3, P(B|A) = 0.6 \neq 0.3 ).
• For any two events A and B, ( P(A \cup B) = P(A) + P(B) - P(A \cap B) ); if disjoint, ( P(A \cap B) = 0 ).
• The conditional probability ( P(A|B) ) is not defined if ( P(B) = 0 ).
• In binomial distribution, the trials must be independent and probability of success constant across trials.
• The sum of probabilities in any probability distribution is always 1.
• verify from NCERT: Exact statement of Theorem of Total Probability in Class 12 Maths textbook (Chapter 13).

Difficulty Level

Intermediate — requires understanding of foundational probability concepts and ability to apply formulas in multi-step problems; direct formula use is easy, but conditional and Bayes’ problems involve logical setup.

Common CUET Traps (3 bullets)

• Trap: Assuming ( P(A|B) = P(B|A) ) implies ( P(A) = P(B) ). Avoid: Use ( P(A|B) = \frac{P(A \cap B)}{P(B)} ) and check actual values; they are equal only if ( P(A) = P(B) ) and ( P(A \cap B) ) satisfies symmetry.
• Trap: Using binomial formula when trials are not independent (e.g., drawing cards without replacement). Avoid: Binomial applies only to with-replacement or large population scenarios; otherwise use hypergeometric or conditional methods.
• Trap: Confusing prior and posterior probabilities in Bayes’ Theorem. Avoid: Prior is ( P(E_i) ), posterior is ( P(E_i|A) ); always label events clearly and follow numerator-denominator structure.

Practice MCQs (5 questions)

Q1. A bag contains 3 red and 4 black balls. Two balls are drawn one after the other with replacement. What is the probability that both are red?
A. ( \frac{6}{49} )
B. ( \frac{9}{49} )
C. ( \frac{3}{7} )
D. ( \frac{2}{7} )
Answer: B
Explanation: With replacement, ( P(\text{both red}) = \frac{3}{7} \times \frac{3}{7} = \frac{9}{49} ).
Why others fail: Option A is ( \frac{^3C_2}{^7C_2} ), which is correct only for without replacement.

Q2. If ( P(A) = 0.5, P(B) = 0.6, P(A \cup B) = 0.8 ), find ( P(A \cap B) ).
A. 0.1
B. 0.2
C. 0.3
D. 0.4
Answer: C
Explanation: Using ( P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow 0.8 = 0.5 + 0.6 - P(A \cap B) \Rightarrow P(A \cap B) = 0.3 ).
Why others fail: Option B (0.2) comes from incorrectly subtracting 0.6 from 0.8.

Q3. A factory has two machines A and B. Machine A produces 60% of items, 2% defective; machine B produces 40%, 1% defective. An item is chosen at random and found defective. What is the probability it was produced by A?
A. ( \frac{3}{4} )
B. ( \frac{2}{3} )
C. ( \frac{1}{2} )
D. ( \frac{1}{3} )
Answer: A
Explanation: By Bayes’ Theorem: ( P(A|D) = \frac{0.6 \times 0.02}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.012}{0.016} = \frac{3}{4} ).
Why others fail: Option B arises from assuming equal contribution or miscalculating denominator.

Q4. A die is thrown 6 times. If “getting an even number” is success, what is the probability of exactly 5 successes?
A. ( \frac{6}{64} )
B. ( \frac{15}{64} )
C. ( \frac{3}{32} )
D. ( \frac{1}{64} )
Answer: C
Explanation: ( n = 6, p = 0.5, x = 5 \Rightarrow P(X=5) = ^6C_5 (0.5)^5 (0.5)^1 = 6 \times \frac{1}{32} \times \frac{1}{2} = \frac{6}{64} = \frac{3}{32} ).
Why others fail: Option A is ( \frac{6}{64} ), which is correct but not simplified — CUET may include both forms; here C is simplified.

Q5. Suppose a random variable X follows binomial distribution with mean 8 and variance 4. Then, the value of n is:
A. 8
B. 12
C. 16
D. 20
Answer: C
Explanation: Mean = np = 8, variance = np(1–p) = 4 ⇒ ( \frac{np(1-p)}{np} = \frac{4}{8} = 0.5 \Rightarrow 1-p = 0.5 \Rightarrow p = 0.5 ), so ( n = \frac{8}{0.5} = 16 ).
Why others fail: Option D (20) comes from incorrectly solving equations or confusing mean and variance.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ ( P(A|B) = \frac{P(A \cap B)}{P(B)} ), only if ( P(B) > 0 ).
• ⚠️ If A and B independent, ( P(A \cap B) = P(A)P(B) ).
• ⚠️ ( P(A \cup B) = P(A) + P(B) - P(A \cap B) ).
• ⚠️ Bayes’ Theorem: ( P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)} ).
• ⚠️ Total Probability: ( P(A) = \sum P(E_i)P(A|E_i) ).
• ⚠️ Bernoulli trial: two outcomes, fixed p, independence.
• ⚠️ Binomial probability: ( P(X = x) = ^nC_x p^x (1-p)^{n-x} ).
• ⚠️ Mean of binomial = np.
• ⚠️ Variance of binomial = np(1–p).
• ⚠️ SD of binomial = ( \sqrt{np(1-p)} ).
• ⚠️ Sum of all probabilities in distribution = 1.
• ⚠️ ( E(X) = \sum x_i p_i ).
• ⚠️ ( Var(X) = E(X^2) - [E(X)]^2 ).
• ⚠️ For fair die, ( E(X) = 3.5 ), ( Var(X) = \frac{35}{12} ).
• ⚠️ If ( P(A \cap B) = P(A)P(B) ), then A and B are independent.
• ⚠️ ( P(A|B) ) undefined if ( P(B) = 0 ).
• ⚠️ Binomial requires fixed n, independent trials, constant p.
• ⚠️ Posterior probability is updated belief after evidence.
• ⚠️ Prior probability is initial belief before evidence.
• ⚠️ verify from NCERT: Example numbers in Chapter 13 of Class 12 Maths for Bayes’ applications.