Q.If P(A)=137, P(B)=139 and P(A∩B)=134, evaluate P(A∣B).
Concept understanding — Conditional Probability
Conditional Probability
Roll a die and ask "what is the chance of an even number?" — that is 3/6. But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at {4,5,6}, and two of those (4 and 6) are even, so the probability becomes 2/3. That change — from the probability of A to the probability of A given that B has already occurred — is conditional probability.
The Idea: Shrink the Sample Space
Conditioning on B throws away every outcome where B is false and treats B as the new "whole world." You measure A only against what is still possible.
Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.
The Definition
For events A and B with P(B)>0,
P(A∣B)=P(B)P(A∩B).
We divide by P(B) to rescale so that B itself has probability 1; the surviving part of A is the overlap A∩B. Checking the die: P(A∩B)=P({4,6})=62 and P(B)=63, so P(A∣B)=3/62/6=32, matching the intuition.
Rearranging gives the multiplication rule P(A∩B)=P(A∣B)P(B), which is usually the easier way to compute a joint probability when a problem says "given that."
Two Cautions
- P(A∣B) and P(B∣A) are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, P(A∣B)=P(B)P(B∣A)P(A).
- If P(A∣B)=P(A), then knowing B tells you nothing about A — the events are independent. That is a special case, not the general rule.
Conditional probability is the foundation of the multiplication theorem, independence, and Bayes' theorem — every "given that" question in this chapter rests on it.
Conditional Probability opens the CBSE Class 12 Probability chapter and is foundational for everything that follows in that unit, including Bayes' theorem and the multiplication rule — making "conditional probability formula class 12 with examples" one of the most searched topics in Class 12 Mathematics. It is equally important for JEE Main and CUET, where conditional probability questions are set almost every year.
Concept: Conditional Probability — the probability of A given B is the ratio of their intersection to the probability of B.
Step 1: Write the formula for conditional probability:
P(A∣B)=P(B)P(A∩B)
Step 2: Substitute the given values:
P(A∣B)=139134
Step 3: Simplify by cancelling 131:
P(A∣B)=94
The value is 94.
Using the definition of conditional probability, P(A∣B)=P(B)P(A∩B). Substituting the given values gives 9/134/13=94.
Conditional probability answers the question: If we know that event B has occurred, how does that change the chance that event A also occurs? The key insight is that knowing B happened restricts the "sample space" to just the outcomes in B. So instead of measuring P(A) against the whole space, we measure P(A∩B) — the part of A that lies inside B — against P(B).
This is exactly the formula:
P(A∣B)=P(B)P(A∩B)
It works because we are renormalising the probability of the overlap by the probability of the new "universe" (B). No extra conditions needed — just plug in the numbers.
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Identify the given probabilities:
P(A)=137, P(B)=139, P(A∩B)=134.
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Write the definition of conditional probability:
P(A∣B)=P(B)P(A∩B)
- Substitute the known values:
P(A∣B)=139134
- Simplify the fraction: The 131 cancels in numerator and denominator, leaving
P(A∣B)=94
A common mistake is to use P(A) instead of P(A∩B) in the numerator. Remember: conditional probability only cares about the part of A that overlaps with B — not the whole of A.
Notice that P(A)=137 was not needed at all for this calculation. Sometimes problems give extra information to test whether you know the correct formula.
The value is 94.
Method: Computing a conditional probability from the three basic quantities
Use this direct approach whenever you are handed P(A), P(B) and P(A∩B) and asked for a conditional probability.
Steps
Step 1: Identify the conditioning event — it sets the denominator.
The event written after the vertical bar is the one you are "given," so its probability goes in the denominator. For P(A∣B) the condition is B.
Step 2: Apply the definition.
P(A∣B)=P(B)P(A∩B).
The numerator is always the joint probability P(A∩B) — the overlap — never P(A) on its own.
Step 3: Substitute and simplify; ignore any unused data.
Put the given fractions in and simplify. Questions often supply an extra value (such as P(A)) that is not needed — recognising that it plays no role is part of the skill, not a sign you missed a step.
Common Mistakes
Mistake 1: Putting P(A) in the numerator instead of P(A∩B).
Why it's wrong: conditional probability measures only the part of A lying inside B, which is P(A∩B). Using 9/137/13 gives 97, a wrong answer. Correct approach: always use P(A∣B)=P(B)P(A∩B)=9/134/13=94.
Mistake 2: Trying to use the unneeded value P(A)=137.
Why it's wrong: the formula needs only P(A∩B) and P(B); the extra datum is a distractor. Correct approach: recognise which quantities the formula actually requires and ignore the rest.
- CA Foundation 2026Set jan-20261 markMCQQ.If in a class, 50% of the student study mathematics and science and 70% of the student study mathematics, then the probability of a student studying science given that he/she is already studying mathematics is (A) 73 (B) 76 (C) 74 (D) 75
›Reveal solutionSolution
Conditional probability P(S∣M)=P(M)P(M∩S).
Step 1 — identify the probabilities
50% study both maths and science, so P(M∩S)=0.5; 70% study maths, so P(M)=0.7.
Step 2 — apply the conditional-probability formula
P(S∣M)=P(M)P(M∩S)=0.70.5=75.
Watch outDivide by the given event's probability: since maths is given, the denominator is P(M)=0.7, not P(S) or the total. Dividing the other way (0.7/0.5) gives a value above 1, which is impossible for a probability.
Tip"Given that" tells you the denominator. Here it is "given studying mathematics," so put P(M) on the bottom: 0.5/0.7=5/7.
✓Final answer(D) 5/7
- CA Foundation 2026Set jan-20261 markMCQQ.If two dice are rolled, then the probability of getting a greater number on the first die than the one on the second, given that the sum should be equal to 7 is (A) 21 (B) 31 (C) 61 (D) 32
›Reveal solutionSolution
Conditional probability on a reduced sample space: P(A∣B)=n(B)n(A∩B).
Step 1 — list the outcomes with sum 7.
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)⇒n(B)=6.
Step 2 — count first die greater than second, among those.
(4,3),(5,2),(6,1) → 3 outcomes.
Step 3 — conditional probability.
P=63=21.
Watch outThe condition "sum = 7" shrinks the sample space to those 6 outcomes — divide by 6, not by the full 36. Using 3/36 gives 1/12, which isn't even an option.
TipNone of the sum-7 pairs are ties, so by symmetry "first > second" and "first < second" split the 6 outcomes evenly — the answer is simply half.
✓Final answer(A) 1/2
- AHSEC Higher Secondary (HS) Final Examination 2026Set ANNUAL1 markMCQQ.If A and B are two events such that A⊂B and P(B)=0, then which of the following is correct?(i) P(A∣B)=P(A)P(B)(ii) P(A∣B)<P(A)(iii) P(A∣B)≥P(A)(iv) None of the above
›Reveal solutionSolution
A⊂B⇒A∩B=A⇒P(A∣B)=P(B)P(A)≥P(A) — option (iii).
By definition P(A∣B)=P(B)P(A∩B).
Because A⊂B, we have A∩B=A, so
P(A∣B)=P(B)P(A).
Since 0<P(B)≤1, the factor P(B)1≥1, hence
P(A∣B)=P(B)P(A)≥P(A).
✓Final answer(iii) P(A∣B)≥P(A).
- CA Foundation 2023Set jun-20231 markMCQQ.If P(A)=31,P(B)=41,P(A/B)=61, the probability P(B/A) is (A) 81 (B) 41 (C) 83 (D) 21
›Reveal solutionSolution
P(B/A) = P(A∩B)/P(A) = (1/24)/(1/3) = 1/8.
Step 1 — Find the joint probability
P(A∩B)=P(A/B)P(B)=61×41=241
Step 2 — Apply the definition of conditional probability
P(B/A)=P(A)P(A∩B)=1/31/24=243=81
Watch outP(A/B) and P(B/A) are not equal — you must recompute the joint probability first, then divide by P(A), not P(B).
TipAnchor everything on P(A∩B): both conditionals flow from it via division by the conditioning event's probability.
✓Final answer(A) 81
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
- CA Foundation 2022Set dec-20221 markMCQQ.If P(A)=31, P(B)=43 and P(A∪B)=1211 then P(AB) is: (A) 61 (B) 94 (C) 21 (D) 81
›Reveal solutionSolution
P(A∩B)=1/6, so P(B|A)=(1/6)/(1/3)=1/2.
Step 1 — Intersection via the addition rule
P(A∩B)=P(A)+P(B)−P(A∪B)=31+43−1211=124+9−11=122=61
Step 2 — Apply the conditional-probability formula
P(AB)=P(A)P(A∩B)=1/31/6=21
Watch outOption (A) 1/6 is just P(A∩B) — you must still divide by P(A) to get the conditional probability.
TipConditional probability always divides the joint probability by the probability of the given (conditioning) event.
✓Final answer(C) 21
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
- CA Foundation 2021Set dec-20211 markMCQQ.For any two dependent events A and B, P(A)=5/9 and P(B)=6/11 and P(A∩B)=10/33. What are the values of P(A/B) and P(B/A)? (A) 5/9, 6/11 (B) 5/6, 6/11 (C) 1/9, 2/9 (D) 2/9, 4/9
›Reveal solutionSolution
Divide the joint probability by the conditioning event's probability: P(A∣B)=5/9, P(B∣A)=6/11.
Step 1 — Apply the conditional probability formula for P(A∣B)
P(A∣B)=P(B)P(A∩B)=6/1110/33=3310×611=198110=95
Step 2 — Apply it for P(B∣A)
P(B∣A)=P(A)P(A∩B)=5/910/33=3310×59=16590=116
Step 3 — Sanity check
Since P(A)P(B)=(5/9)(6/11)=10/33=P(A∩B), the conditionals collapse to the marginals — consistent with the computed values.
Watch outThe trap is to multiply P(A∩B) by P(B) instead of dividing (giving small fractions like 1/9, 2/9 in options C/D). Always divide by the given/conditioning event.
TipP(A∣B) = joint over the second letter's probability; P(B∣A) = joint over the first letter's probability.
✓Final answer(A) 5/9, 6/11
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
- CA Foundation 2021Set dec-20211 markMCQQ.In a group of 20 males and 15 females, 12 males and 8 females are service holders. What is the probability that a person selected at random from the group is a service holder given that the selected person is a male? (A) 0.40 (B) 0.60 (C) 0.45 (D) 0.55
›Reveal solutionSolution
Condition on males only: 12 service holders out of 20 males = 0.60.
Step 1 — Identify the reduced sample space
Given the person is male, only the 20 males matter.
Step 2 — Apply the conditional formula
P(service∣male)=total malesmale service holders=2012=0.60
Watch outDo not divide by the full group of 35 — the condition 'given male' shrinks the denominator to 20.
Tip'Given that ...' problems: throw away everyone outside the given category, then take the simple fraction.
✓Final answer(B) 0.60
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
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