ISC Class 12 Maths Probability – Conditional Probability Explained

Ever wondered how likely something is *after* something else has happened?

💡 In Simple Words: Conditional probability tells us the chance of an event when we already know another event occurred. Think of it like guessing the colour of a sock you’ll pick after you’ve already taken out a red one.

What is conditional probability?

In everyday language, an event is just a thing that can happen – like rolling a 4 on a dice or picking a queen from a deck. Conditional probability asks: "If we know that event B has happened, what’s the chance that event A also happens?" We write it as P(A|B) – the vertical bar reads "given".

Formula you’ll need

The math looks like this:

P(A|B) = \frac{P(A\cap B)}{P(B)}

Here P(A\cap B) means the probability that *both* A and B happen together (the intersection). P(B) is the probability of the known event B. The fraction simply says: "Take the chance of both happening, then shrink it by the chance that B happened in the first place."

How to compute it – step by step

Worked example 1: Drawing cards

Suppose you pick one card from a standard 52‑card deck. Let A be "the card is a king" and B be "the card is a face card (jack, queen or king)." First, how many face cards are there? There are 12 (4 jacks, 4 queens, 4 kings). So P(B)=12/52=3/13.

How many cards are both a king *and* a face card? That’s just the 4 kings, so P(A∩B)=4/52=1/13.

Now plug into the formula:

P(A|B)= (1/13) ÷ (3/13) = 1/3. So, given that you already know the card is a face card, the chance it’s a king is one‑third.

Worked example 2: Rolling dice

Imagine you roll two fair six‑sided dice. Let A be "the sum of the dice is 7" and B be "the first die shows 4." First, P(B)=1/6 because any one face is equally likely.

What’s P(A∩B)? If the first die is 4, the second must be 3 to make a total of 7. That single outcome out of 36 possible pairs gives P(A∩B)=1/36.

Thus P(A|B)= (1/36) ÷ (1/6) = 1/6. Knowing the first die is 4 leaves six equally likely outcomes for the second die, only one of which makes the sum 7.

Common pitfalls

• Forgetting that P(B) must be >0. If B can never happen, the conditional probability is undefined.
• Mixing up P(A|B) with P(B|A). They are generally different unless the events are independent.
• Counting the intersection incorrectly – remember to count only the outcomes that satisfy *both* conditions.

Quick comparison

📝 Likely Exam Questions

• Q1. A bag contains 5 red and 7 blue balls. Two balls are drawn without replacement. Find the probability that the second ball is red given that the first ball drawn was blue.
  Answer: P(R₂|B₁)= (5/12) ÷ (7/12) = 5/7.
• Q2. From a deck of 52 cards, what is the probability that a drawn card is a queen given that it is a face card?
  Answer: P(Q|Face)= (4/52) ÷ (12/52) = 1/3.
• Q3. Two dice are rolled. Find the probability that the sum is 9 given that at least one die shows a 3.
  Answer: Possible outcomes with a 3: (3,1‑6) and (1‑6,3) → 11 outcomes (excluding double‑count of (3,3)). Those giving sum 9 are (3,6) and (6,3). So P=2/11.
• Q4. Explain why P(A|B)=P(A) when A and B are independent.
  Answer: Independence means the occurrence of B does not affect A, so the intersection equals the product: P(A∩B)=P(A)P(B). Substituting gives P(A|B)=P(A)P(B)/P(B)=P(A).