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IIT Medico Guide In collaboration with Ashoka Group of Schools

Maths · Class 12 · Chapter 13

Probability

Class 11 probability counted outcomes. This chapter asks how a probability changes once we know something has happened, which leads to conditional probability, independence and Bayes' theorem. It ends with random variables, their mean and variance, and the binomial distribution.

In this chapter: conditional probability and its properties, the multiplication rule, independent events, the theorem of total probability, Bayes' theorem, random variables and probability distributions, mean and variance of a random variable, and the binomial distribution.

Conditional probability

If E and F are events of the same sample space and P(F) ≠ 0, the probability of E given that F has occurred is

P(E | F) = P(E ∩ F)P(F)F becomes the new, reduced sample space

For equally likely outcomes this is simply n(E ∩ F)/n(F). Properties: P(S | F) = P(F | F) = 1; P(E′ | F) = 1 − P(E | F); and P(A ∪ B | F) = P(A | F) + P(B | F) − P(A ∩ B | F).

Worked example: A family has two children. Given that at least one is a boy, find the probability that both are boys.
Solution: S = {BB, BG, GB, GG}. F = at least one boy = {BB, BG, GB}, and E ∩ F = {BB}. So P(E | F) = 1/3, not 1/2. The information "at least one boy" removes only GG, leaving three equally likely cases.

Multiplication rule and independence

P(E ∩ F) = P(E) P(F | E) = P(F) P(E | F)P(E ∩ F ∩ G) = P(E) P(F | E) P(G | E ∩ F)

E and F are independent if the occurrence of one does not change the probability of the other, that is, P(E ∩ F) = P(E) P(F). Equivalently P(E | F) = P(E) when P(F) ≠ 0.

  • If E and F are independent, so are E and F′, E′ and F, and E′ and F′. Then, for example, P(E ∪ F) = 1 − P(E′) P(F′).
  • Three events are mutually independent only if every pair is independent and P(E ∩ F ∩ G) = P(E) P(F) P(G).
  • Independent is not the same as mutually exclusive. Mutually exclusive events with non-zero probabilities are never independent, because P(E ∩ F) = 0 while P(E) P(F) > 0.

Total probability and Bayes' theorem

Let E1, E2, ..., En be a partition of the sample space (pairwise disjoint, together covering S, each with non-zero probability), and let A be any event.

P(A) = ∑ P(Ej) P(A | Ej) (total probability)P(Ei | A) = P(Ei) P(A | Ei)∑ P(Ej) P(A | Ej) (Bayes' theorem)

The Ei are called hypotheses, P(Ei) the prior probabilities, and P(Ei | A) the posterior probabilities. Bayes' theorem works "backwards": we see the result A and ask which cause probably produced it. A tree diagram keeps the numbers organised.

Probability tree for Bayes theoremwww.iitmedicoguide.com1/21/23/52/51/43/4startBag IBag IIRed→ 3/10Black→ 1/5Red→ 1/8Black→ 3/8P(Red) = 3/10 + 1/8 = 17/40P(Bag I | Red) = (3/10) ÷ (17/40) = 12/17www.iitmedicoguide.com
Bag I holds 3 red and 2 black balls; Bag II holds 1 red and 3 black. Multiply along each path; the red paths add to P(Red), and Bayes' theorem takes the fraction of that total coming from Bag I.
Worked example: Bag I contains 3 red and 2 black balls; Bag II contains 1 red and 3 black balls. A bag is chosen at random and a ball drawn from it is red. Find the probability that it came from Bag I.
Solution: P(E1) = P(E2) = 1/2, P(R | E1) = 3/5, P(R | E2) = 1/4. Total probability: P(R) = (1/2)(3/5) + (1/2)(1/4) = 3/10 + 1/8 = 17/40. Bayes: P(E1 | R) = (3/10) ÷ (17/40) = 12/17.

Random variables and their distributions

A random variable X is a real-valued function on the sample space. Its probability distribution lists each value xi with its probability pi, where every pi ≥ 0 and ∑ pi = 1. The second condition is how "find k" questions are solved.

Mean: μ = E(X) = ∑ xi piVariance: Var(X) = E(X²) − [E(X)]² = ∑ xi² pi − μ²Standard deviation: σ = √Var(X)
Worked example: Two fair coins are tossed and X is the number of heads. Find the mean and variance of X.
Solution: X takes the values 0, 1, 2 with probabilities 1/4, 1/2, 1/4 (they add to 1). E(X) = 0 + 1/2 + 2/4 = 1. E(X²) = 0 + 1/2 + 4/4 = 3/2. Var(X) = 3/2 − 1 = 1/2, and the mean is 1.

Bernoulli trials and the binomial distribution

Bernoulli trials and the binomial distribution were removed from the rationalised NCERT book, but they are still part of the MHT‑CET syllabus. Trials are Bernoulli trials if there is a fixed number n of them, they are independent, each has exactly two outcomes (success or failure), and the probability of success p is the same in every trial. With q = 1 − p, the number of successes X has the binomial distribution B(n, p):

P(X = r) = nCr pr qn − r, r = 0, 1, ..., nmean = np, variance = npq
Worked example: A fair die is thrown 4 times. Find the probability of getting exactly two sixes.
Solution: n = 4, p = 1/6, q = 5/6. P(X = 2) = 4C2 (1/6)² (5/6)² = 6 × 25/1296 = 25/216.

"At least one" questions are fastest by the complement: P(X ≥ 1) = 1 − qn.

Common mistakes: (1) Confusing P(E | F) with P(F | E); in Bayes questions, write down clearly which event is known. (2) Treating mutually exclusive events as independent. (3) Using P(A ∪ B) = P(A) + P(B) for events that can happen together. (4) Forgetting that the variance formula subtracts the square of the mean, not the mean itself. (5) Applying the binomial formula when trials are not independent, such as drawing without replacement.

JEE and MHT‑CET focus

  • Conditional probability with dice, coins, cards and families, using the reduced sample space.
  • Testing independence and using P(E ∪ F) = 1 − P(E′)P(F′) for independent events.
  • Bayes' theorem with two or three hypotheses (bags, machines, tests, truthful witnesses).
  • Finding k in a probability distribution, then its mean and variance.
  • Binomial distribution: probabilities, and finding n and p from given mean and variance.

Practice questions

If P(A) = 0.8, P(B) = 0.5 and P(B | A) = 0.4, then P(A | B) is:

  1. 0.32
  2. 0.64
  3. 0.16
  4. 0.25
Show answer
B. P(A ∩ B) = 0.8 × 0.4 = 0.32; P(A | B) = 0.32/0.5 = 0.64.

A and B are independent with P(A) = 0.3 and P(B) = 0.4. P(A ∪ B) is:

  1. 0.7
  2. 0.12
  3. 0.58
  4. 0.42
Show answer
C. 0.3 + 0.4 − 0.3 × 0.4 = 0.58.

If P(E) = 1/2, P(F) = 1/3 and P(E ∩ F) = 1/6, then E and F are:

  1. Mutually exclusive
  2. Independent
  3. Exhaustive
  4. Complementary
Show answer
B. P(E)P(F) = 1/6 = P(E ∩ F). They are not mutually exclusive, since P(E ∩ F) ≠ 0.

Two events with non-zero probabilities that are mutually exclusive are:

  1. Always independent
  2. Never independent
  3. Independent only if equally likely
  4. Independent only if exhaustive
Show answer
B. P(E ∩ F) = 0 but P(E)P(F) > 0.

Machine A makes 60% of a factory's bolts and machine B makes 40%. Their defect rates are 2% and 3%. A bolt picked at random is defective. The probability that it was made by A is:

  1. 2/5
  2. 1/2
  3. 3/5
  4. 1/3
Show answer
B. P(A)P(D | A) = 0.6 × 0.02 = 0.012 and P(B)P(D | B) = 0.4 × 0.03 = 0.012, so the answer is 0.012/0.024.

The mean of the number obtained on throwing a fair die is:

  1. 3
  2. 3.5
  3. 4
  4. 21
Show answer
B. (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5.

A binomial distribution has mean 4 and variance 3. The number of trials n is:

  1. 12
  2. 16
  3. 8
  4. 7
Show answer
B. npq/np = q = 3/4, so p = 1/4 and n = 4/(1/4) = 16.

A random variable X takes values 0, 1, 2 with probabilities k, 2k, 3k. The value of k is:

  1. 1/3
  2. 1/6
  3. 1/5
  4. 1/4
Show answer
B. k + 2k + 3k = 1 gives k = 1/6.
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