Q.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.
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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). …
Concept: Conditional Probability (Bayes' Theorem)
Let E1 be the event that the die shows a six, and E2 be the event that it does not show a six.
Let A be the event that the man reports a six.
Step 1: Prior probabilities
P(E1)=61, P(E2)=65.
Step 2: Likelihoods (probability of reporting a six given the actual outcome)
If it is a six, he tells truth: P(A∣E1)=43.
If it is not a six, he lies and reports a six: P(A∣E2)=41.
Step 3: Apply Bayes' Theorem …
A Bayes' theorem problem. The probability the die actually shows a six, given the report, is 83.
Let S be the event "the die shows a six" and R the event "the man reports a six".
P(S)=61,P(S′)=65.
He speaks truth 3 of 4 times, so if it is a six he correctly reports six with probability 43, and if it is not a six he (lying) reports six with probability 41:
P(R∣S)=43,P(R∣S′)=41.
By Bayes' theorem, …
Method: Bayes' Theorem (finding the cause from the observed outcome)
Use this "reliability of a witness" pattern: someone reports an outcome and you want the probability the outcome is genuinely what they claim. You know how truthful the reporter is; you want P(event true∣event reported).
Steps
Step 1: Set up the partition of possible causes.
List the mutually exclusive, exhaustive hypotheses (here: the die actually shows a six, or it does not) and write their prior probabilities P(Ei) (these must sum to 1).
Step 2: Write each likelihood.
For each cause, state P(A∣Ei) — the probability of the observed outcome (here: the man reports a six) under that cause.
Step 3: Get the total probability of the outcome (the denominator).
By the law of total probability,
P(A)=∑iP(Ei)P(A∣Ei).
Step 4: Apply Bayes' theorem for the cause you want. …
Common Mistakes
Mistake 1: Answering 43 (his truth-telling rate).
Why it's wrong: 43 is P(reports six∣it is a six), whereas the question asks the reverse, P(it is a six∣reports six). Correct approach: apply Bayes' theorem.
Mistake 2: Forgetting the "lie" branch in the denominator. …
- 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. …
- 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. …
- Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQQ.If P(A∩B)=70% and P(B)=85%, then P(A/B)=(a) 1417(b) 1714(c) 87(d) 81
›Reveal solutionSolution
Conditional probability is just P(A∣B)=P(B)P(A∩B) — plug in the given values.
Given P(A∩B)=70%=0.70 and P(B)=85%=0.85: …
- Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markQ.A ludo die is rolled. If the outcome is an odd number, what is the probability that it is a prime number?
›Reveal solutionSolution
A ludo/standard die shows 1–6; restrict the sample space to the given condition (odd) and count how many of those are prime.
A die has outcomes {1,2,3,4,5,6}. Given the outcome is odd, the reduced sample space is
{1,3,5}(3 equally likely outcomes).
…
- Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 markQ.If P(A)=137, P(B)=139, and P(A∩B)=134, find the value of P(A/B).
›Reveal solutionSolution
Conditional probability P(A/B)=P(B)P(A∩B).
Given P(A)=137, P(B)=139, P(A∩B)=134.
…
- 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). …
- 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. …
- 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. …
- 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. …
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