Ever wondered why the chance of pulling a red marble changes after you already picked a blue one? That's conditional probability in action.
Think of it like this: you want to know the chance of something happening, but you already know another event occurred. It’s just updating your odds based on new information.
What is Conditional Probability?
Conditional probability measures the likelihood of an event A happening given that another event B has already happened. The word “given” is the key – it tells you you’re not starting from scratch.
Formula and How to Use It
The standard formula is:
P(A|B) = P(A ∩ B) / P(B)
Here P(A|B) means “the probability of A given B”. The symbol “∩” stands for “and” – the chance that both A and B happen together. And P(B) is the chance of the known event B.
Imagine water flowing through two pipes that join. The total flow through the joint (A ∩ B) divided by the flow in the second pipe (B) tells you what fraction of water in pipe B also goes through pipe A. That’s exactly what the formula does with chances.
Worked Example 1: Drawing Cards
Suppose you draw one card from a standard deck of 52 cards. What’s the probability that the card is a king given that it’s a face card?
- Step 1: Identify B – “face card”. There are 12 face cards (J, Q, K of each suit). So P(B) = 12/52 = 3/13.
- Step 2: Identify A – “king”. There are 4 kings.
- Step 3: Find A ∩ B – cards that are both kings and face cards. All kings are face cards, so there are 4 such cards. P(A ∩ B) = 4/52 = 1/13.
- Step 4: Plug into the formula: P(A|B) = (1/13) / (3/13) = 1/3. So once you know the card is a face card, the chance it’s a king jumps from 4/52 to 1/3.
Worked Example 2: Selecting Students
In a class of 30 students, 18 study maths, 12 study physics, and 6 study both. Find the probability that a randomly chosen student studies physics given they study maths.
- Event B = “studies maths”. P(B) = 18/30 = 3/5.
- Event A = “studies physics”.
- A ∩ B = “studies both maths and physics”. That’s 6 students, so P(A ∩ B) = 6/30 = 1/5.
- Now P(A|B) = (1/5) / (3/5) = 1/3. One out of three maths students also studies physics.
Common Mistakes to Avoid
- Forgetting to check that P(B) ≠ 0. If B never happens, the conditional probability is undefined.
- Mixing up P(A|B) with P(B|A). The order matters – “given B” is not the same as “given A”.
- Using the total number of outcomes instead of the reduced sample space after B occurs.
Quick Comparison Table
| Aspect | Unconditional Probability | Conditional Probability |
|---|---|---|
| Definition | Chance of an event without extra info. | Chance of an event after another event is known. |
| Formula | P(A) = favourable / total | P(A|B) = P(A∩B) / P(B) |
| Sample Space | All possible outcomes. | Only outcomes where B occurs. |
| Typical Question | “What’s the chance of a 6 on a die?” | “What’s the chance of a 6 on a die given the roll is even?” |
📝 Likely Exam Questions
- In a bag there are 5 red, 4 blue and 3 green balls. One ball is drawn without replacement. Find the probability that the second ball is blue given that the first ball drawn was red.
Answer: After a red ball is removed, 11 balls remain with 4 blue. So P = 4/11. - A box contains 6 defective and 14 good bulbs. Two bulbs are drawn one after the other without replacement. What is the probability that the second bulb is good given that the first bulb drawn was defective?
Answer: After removing a defective bulb, 19 bulbs remain with 14 good. So P = 14/19. - From a deck of 52 cards, find the probability of drawing a queen given that the drawn card is a face card.
Answer: P(queen|face) = (4/52) / (12/52) = 4/12 = 1/3. - In a survey, 40% like tea, 30% like coffee, and 20% like both. What is the probability that a person likes coffee given they like tea?
Answer: P(coffee|tea) = 0.20 / 0.40 = 0.5. - Explain why P(A|B) = P(A) when events A and B are independent.
Answer: Independence means the occurrence of B does not affect A, so P(A∩B)=P(A)P(B). Plugging into the formula gives P(A|B)=P(A)P(B)/P(B)=P(A).