Conditional probability and independent events

Objectives and prerequisites

Compute a conditional probability by restricting and renormalizing the sample space; distinguish conditional, joint and marginal probabilities; use the multiplication rule with or without replacement; and test independence instead of assuming it. Prerequisites: sample spaces, event intersections, complements and basic counting. Conditional probability is explicitly listed in Section 1 of the GATE 2026 CS syllabus. Pairwise versus mutual independence is included as a useful distinction, not a claim of separate syllabus wording.

1. Conditioning changes the reference population

For P(B)>0, P(A|B)=P(A∩B)/P(B). Read this as “the probability of A given that B has occurred.” B is the information already known. We discard outcomes outside B, then rescale the probabilities inside B to sum to one. In a finite equally likely model, this becomes |A∩B|/|B|. The denominator is the size or probability of the conditioning event, not the original universe.

Example: roll a fair die. Given that the result is even, find the probability it exceeds three. The restricted set is B={2,4,6}; successful outcomes are {4,6}. Therefore P(A|B)=2/3. Without the condition, P(A)=3/6=1/2. The joint probability P(A∩B)=2/6=1/3 is a third, different number.

Reversing the condition asks a different question. For this die example P(B|A) also happens to be 2/3 because P(A)=P(B), but this is accidental. If A is “six” and B is “even”, then P(A|B)=1/3 while P(B|A)=1. Every six is even; not every even result is six. Always write the event after the vertical bar first in your interpretation.

The elementary ratio is undefined when P(B)=0. Do not assign zero to 0/0. Conditioning on a continuous variable taking an exact value uses a more advanced construction, not this event-ratio calculation.

2. Tables keep joint and conditional data separate

Among 100 equally likely records, 40 use system A, 60 use system B. Eight A records and twelve B records have an error. There are 20 errors overall. P(error and A)=8/100=.08, P(error|A)=8/40=.20, and P(A|error)=8/20=.40. The same cell count eight has different denominators for different questions.

To solve from a table, reconstruct row totals, column totals and the grand total. For P(row event|column event), divide the intersection cell by the conditioning column total. Percentages must be attached to their populations: “20% of A records” is conditional, not 20% of all records.

Complementation works inside a fixed condition: P(Aᶜ|B)=1−P(A|B). But P(A|Bᶜ) is not generally 1−P(A|B); it changes the population. In the table above both system groups have error rate .20, so changing the system condition does not complement the rate.

3. Multiplication rule and sequential sampling

Rearranging the definition gives P(A∩B)=P(A)P(B|A) whenever P(A)>0. Also P(A∩B)=P(B)P(A|B) whenever P(B)>0. These rules do not require independence. For three events with the needed positive conditioning probabilities,

P(A∩B∩C)=P(A)P(B|A)P(C|A∩B).

Example: a bag has three red and two blue balls. Draw twice uniformly without replacement. P(first red and second red)=(3/5)(2/4)=3/10. After a red first draw, only two red balls among four remain. Multiplying (3/5)² would ignore this change. The second draw still has unconditional red probability 3/5 by symmetry, but its probability conditional on a red first draw is 1/2. Marginal and conditional probabilities need not match.

Exactly one red can occur in two disjoint orders: red-blue has probability (3/5)(2/4)=3/10, and blue-red has probability (2/5)(3/4)=3/10. Add to obtain 3/5. With replacement and independent random draws, red-red instead has probability (3/5)²=9/25. The physical process determines the model.

4. Independence is a product identity

Events A and B are independent when P(A∩B)=P(A)P(B). When P(B)>0, this is equivalent to P(A|B)=P(A). Independence means learning B does not change the probability of A. It is symmetric. The product definition remains valid even if one event has probability zero, whereas a conditional expression with a zero denominator does not.

Example: for two independent fair coin tosses, let A be “first toss is H” and B be “second toss is H.” P(A)=P(B)=1/2 and P(A∩B)=1/4; the product test passes. On one fair die, let C={1,2} and D={2,3,4}. P(C∩D)=1/6 and P(C)P(D)=(1/3)(1/2)=1/6, so these overlapping events are also independent. Independence does not mean “no shared outcomes.”

Disjoint events with positive probabilities cannot be independent: their intersection probability is zero but their probability product is positive. Events “die is one” and “die is two” are disjoint and dependent. If independent A,B have positive probabilities, they must be able to occur together with positive probability.

Independence also passes to complements: P(A∩Bᶜ)=P(A)−P(A∩B)=P(A)(1−P(B)). Thus A and Bᶜ are independent; similarly Aᶜ and B, and Aᶜ and Bᶜ. This is a deduction from the product test, not a new assumption.

5. Pairwise and mutual independence

For three events, pairwise independence checks each pair. Mutual independence additionally requires P(A∩B∩C)=P(A)P(B)P(C). Pairwise checks alone do not establish the triple identity.

Use two independent fair coin tosses: A means the first is H, B the second is H, and C the two tosses agree. Each event has probability 1/2. Every pair intersects only at HH, with probability 1/4, so every pair is independent. But the triple intersection is also {HH}, with probability 1/4, not (1/2)³=1/8. They are not mutually independent. C is determined once both toss results are known.

For more events, mutual independence requires the product rule for every subcollection of at least two events. When a problem says all trials are independent, it normally grants the joint independence needed to multiply across trials. Do not infer it from only pairwise numerical checks.

Solving checklist

Name A and the known condition B separately. Check P(B)>0. Find the joint event before dividing. In sequential draws, update the available counts after each observed outcome. Test independence with a joint probability, or use an explicit modeling assumption stated in the question. Never confuse independent with disjoint; never reverse a condition without recalculating. If the final answer refers to a changed population, make that denominator visible.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CP01. A fair die is rolled. Given an even result, what is the probability that it exceeds three?

A. 1/2 B. 1/3 C. 2/3 D. 1

Answer: C

A: This ignores the condition.

B: This is the joint probability on the original six-outcome space.

C: Among 2,4,6, the two values 4,6 exceed three.

D: The even value 2 does not exceed three.

CP02. A is “six” and B is “even” on a fair die. What is P(B|A)?

A. 1 B. 1/3 C. 1/2 D. 1/6

Answer: A

A: Given six, evenness is certain.

B: This reverses the condition and gives P(A|B).

C: This is the unconditional P(B).

D: This is P(A∩B), before conditioning.

CP03. Draw twice uniformly without replacement from three red and two blue balls. What is P(both red)?

A. 9/25 B. 3/5 C. 1/2 D. 3/10

Answer: D

A: This would apply to independent draws with replacement.

B: This is the probability of red on the first draw only.

C: This is the probability of second red conditional on first red.

D: Multiply 3/5 by 2/4 to obtain 3/10.

CP04. P(A)=0.4, P(B)=0.5, P(A∩B)=0.2. Which statement is correct?

A. A and B are disjoint B. A and B are independent C. P(A|B)=0.2 D. P(A∪B)=0.9

Answer: B

A: Their intersection has positive probability, so they are not disjoint.

B: The intersection 0.2 equals 0.4×0.5.

C: Divide 0.2 by 0.5 to get 0.4.

D: The union is 0.4+0.5−0.2=0.7.

CP05. Two disjoint events each have strictly positive probability. They are:

A. always independent B. always dependent C. necessarily complements D. necessarily equally likely

Answer: B

A: Their zero intersection cannot equal a strictly positive product.

B: The product identity for independence fails.

C: They need not cover the sample space.

D: Disjointness imposes no equality on their probabilities.

CP06. For two independent fair tosses, A=first H, B=second H, C=matching tosses. Which is correct?

A. The three are mutually independent B. No pair is independent C. They are pairwise independent but not mutually independent D. P(A∩B∩C)=1/8

Answer: C

A: The triple intersection is 1/4 rather than the product 1/8.

B: Every pair has intersection 1/4 and marginal product 1/4.

C: All pair tests pass; the triple test fails.

D: HH is the triple intersection and has probability 1/4.

Source and scope

This lesson develops conditional probability in Section 1 of the official GATE 2026 CS syllabus and the independence concepts needed to use it correctly. Total probability and Bayes theorem follow separately. Original teaching and practice; not full-syllabus coverage.

Official GATE 2026 CS syllabus (verified 2026-10-14).