Quantitative Aptitude · Probability

Conditional Probability for CA Foundation: Formula, Method and Solved MCQs

Updated 1 October 2026 · Fact-checked

Conditional probability is the chance of event A happening when you already know event B has occurred. Use P(A | B) = P(A ∩ B) ÷ P(B), with P(B) > 0. In equally likely problems, shrink the sample space to B and count how many outcomes also belong to A.

Understand Conditional Probability

Probability changes when you get new information. The chance that a card is a king is 4/52. If you are told the card is a face card, the chance becomes 4/12. That updated chance is a conditional probability.

We write it as P(A | B). Read it as "probability of A given B". The event after the bar is the condition. It has already happened, so it becomes your new sample space.

Think of it as zooming in. Out of all outcomes, you look only at those in B. Then you ask what fraction of those are also in A. That is why the formula divides P(A ∩ B) by P(B), not by 1.

This matters most for dependent events. Drawing cards or balls without replacement is the classic case. The first draw changes what is left, so the second probability changes. If knowing B does not change the chance of A, the events are independent.

P(A | B) and P(B | A) are usually different. Always check which event is the condition before you calculate.

Key formulas to remember

Conditional probability: P(A | B) = P(A ∩ B) ÷ P(B); Valid only when P(B) > 0. B is the given (condition) event.
Reverse conditional: P(B | A) = P(A ∩ B) ÷ P(A); Valid only when P(A) > 0. Not the same as P(A | B) in general.
Multiplication rule: P(A ∩ B) = P(B) × P(A | B) = P(A) × P(B | A); Use it for draws without replacement. Multiply the first probability by the updated second one.
Equally likely outcomes: P(A | B) = n(A ∩ B) ÷ n(B); Count only inside B. Works when all outcomes are equally likely.
Complement given B: P(A' | B) = 1 − P(A | B); The condition B must stay the same on both sides.
Test for independence: A and B are independent if P(A | B) = P(A), that is, P(A ∩ B) = P(A) × P(B); If P(A | B) differs from P(A), the events are dependent.

How to solve Conditional Probability questions

Use this method for any conditional probability question. It keeps the condition and the target clear.

1Read the question and find the words "given that", "if it is known that" or "after". The event they describe is B, the condition.
2Name the target event as A. Write the required value as P(A | B).
3List what the question gives you: P(A), P(B), P(A ∩ B), or the counts of outcomes.
4If the outcomes are equally likely, shrink the sample space to B. Count n(B), then count how many of those outcomes are also in A, which is n(A ∩ B).
5Otherwise use P(A | B) = P(A ∩ B) ÷ P(B). If P(A ∩ B) is missing, find it with the multiplication rule.
6For draws without replacement, reduce the totals after each draw. Remove the item drawn from both the favourable count and the total.
7Simplify the fraction and check that the answer lies between 0 and 1.
8Compare with the options and confirm that you did not give P(B | A) by mistake.

Quickest way: Shrink the sample space

When to use it: Use it when outcomes are equally likely, as in dice, cards, coins and balls in a bag. It is usually faster than the formula.

Underline the "given" event. Count its outcomes. This is your new denominator.
From those outcomes only, count the ones that also satisfy the target. This is your numerator.
Write numerator ÷ denominator and simplify.
For two draws without replacement, the second probability is just a fresh fraction with one item removed.
Remove options that are bigger than 1 or equal to the unconditional probability when the condition clearly changes things.
If the counts are not clear in about a minute, skip the question. A wrong answer costs 0.25 marks.

Common mistakes in Conditional Probability

Dividing by the wrong probability, for example computing P(A ∩ B) ÷ P(A) for P(A | B).
Students do not identify which event is the condition.
Fix: The event after the bar is the denominator. Mark the "given" event before you start.
Treating P(A | B) as equal to P(B | A).
The two look similar, and the intersection in the numerator is the same.
Fix: The denominators differ. Check which event has been given and divide by its probability.
Using the same total for the second draw when there is no replacement.
Students copy the first fraction without updating the counts.
Fix: After each draw, reduce the total by 1. Reduce the favourable count by 1 only if the drawn item was of that type.
Counting outside B when using the shrunken sample space.
Students count favourable outcomes from the whole sample space.
Fix: Count only outcomes that are in both A and B. Anything outside B is ignored.
Assuming events are independent without checking.
Students multiply P(A) and P(B) out of habit.
Fix: Multiply P(A) × P(B) only if the question says independent, or if draws are with replacement. Otherwise use P(B) × P(A | B).
Using P(A | B) when P(B) = 0.
Students apply the formula without checking the condition.
Fix: The formula needs P(B) > 0. The condition must be a possible event.

Worked examples

Example 1

A bag has 5 red and 3 blue balls. Two balls are drawn one after another without replacement. Given that the first ball is red, the probability that the second is red is: (a) 5/8 (b) 4/7 (c) 3/7 (d) 1/2

Show the solution

The condition B: the first ball is red. The target A: the second ball is red.
After a red ball is removed, the bag has 4 red and 3 blue balls.
The total is now 7 balls, of which 4 are red.
P(second red | first red) = 4/7.

Answer: (b) 4/7

Example 2

For two events A and B, P(A) = 0.5, P(B) = 0.6 and P(A ∩ B) = 0.24. Find P(B | A). Options: (a) 0.24 (b) 0.30 (c) 0.40 (d) 0.48

Show the solution

P(B | A) has A as the condition, so the denominator is P(A).
P(B | A) = P(A ∩ B) ÷ P(A) = 0.24 ÷ 0.5.
0.24 ÷ 0.5 = 0.48.
Check: 0.48 is not equal to P(B) = 0.6, so A and B are dependent. Option (c) 0.40 is P(A | B) = 0.24 ÷ 0.6, which is the trap.

Answer: (d) 0.48

Example 3

Two fair dice are thrown. Given that the sum is 8, the probability that both dice show even numbers is: (a) 1/12 (b) 1/4 (c) 3/5 (d) 5/36

Show the solution

The condition B: the sum is 8. The outcomes are (2,6), (3,5), (4,4), (5,3), (6,2). So n(B) = 5.
The target A: both numbers even. Among these five outcomes, (2,6), (4,4) and (6,2) qualify. So n(A ∩ B) = 3.
P(A | B) = 3 ÷ 5.
Check with the formula: P(A ∩ B) = 3/36 and P(B) = 5/36. Dividing gives 3/5.

Answer: (c) 3/5

Exam tips

Spot the words "given that", "if it is known" and "after". They tell you the denominator.
For equally likely outcomes, use the shrunken sample space. It saves time and reduces errors.
Options often include P(A ∩ B) and P(B | A) as traps. Check that your answer matches the conditional you were asked for.
For two draws without replacement, update the totals before writing the second fraction.
Long multi-step problems can wait. Answer the quick ones first, since wrong answers lose 0.25 marks each.

Practice questions from Probability

Conditional Probability: frequently asked questions

What is the formula for conditional probability in CA Foundation?

P(A | B) = P(A ∩ B) ÷ P(B), provided P(B) > 0. Here B is the event that is given. For equally likely outcomes you can also use n(A ∩ B) ÷ n(B).

How is conditional probability different from ordinary probability?

Ordinary probability looks at the whole sample space. Conditional probability looks only at the outcomes where the given event has occurred. The denominator therefore becomes P(B) instead of 1.

How do I know if two events are independent?

Events A and B are independent if P(A | B) = P(A), or equivalently P(A ∩ B) = P(A) × P(B). If knowing B changes the chance of A, the events are dependent. Draws without replacement are usually dependent.

Is P(A | B) the same as P(B | A)?

No, not in general. They share the same numerator P(A ∩ B) but have different denominators. They are equal only when P(A) = P(B).