Q.Given that E and F are events such that $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E|F)$ and $P(F|E)$.

Concept: Conditional Probability — the probability of one event given that another has occurred.

Step 1: Recall the formula for conditional probability:

Step 2: Substitute the given values:

Step 3: Similarly,

The idea behind conditional probability is simple: when we say "probability of E given F," we are no longer looking at the whole world of possibilities — we are only considering those outcomes where F has already happened. So the new "universe" is F itself, and within that universe, we want the fraction where E also occurs. That fraction is just the portion of F that overlaps with E, divided by the total size of F.

This is why the formula is:

The numerator is the overlap (both events happen), and the denominator is the condition we are given.

Now let's plug in the numbers.

1. Find $P(E|F)$ We have $P(E \cap F) = 0.2$ and $P(F) = 0.3$. So

2. Find $P(F|E)$ Here the condition is E, so denominator is $P(E) = 0.6$.

Notice that $P(E|F)$ and $P(F|E)$ are not the same, and they don't have to be. Here, knowing that F occurred makes E more likely (2/3 vs 0.6), while knowing that E occurred makes F less likely (1/3 vs 0.3). That makes sense because E is larger than F, so F occupies a smaller fraction of E than E does of F.

Method: Computing P(A|B) and P(B|A) from Joint and Marginal Probabilities

Use this when a problem gives $P(A)$, $P(B)$ and $P(A\cap B)$ and asks for one or both conditional probabilities.

Steps

Step 1: Identify the conditioning event — it goes in the denominator.

The event written after the bar is what is assumed to have happened, so we measure against it:

Step 2: Swap roles for the reverse conditional.

The numerator (the joint overlap) is the same both times; only the denominator changes.

Step 3: Interpret the asymmetry.

$P(A\mid B)\ne P(B\mid A)$ in general. Whichever conditioning event is smaller gives the larger conditional probability, because the shared overlap is a bigger fraction of a smaller set.

Common Mistakes

Mistake 1: Swapping the denominators of $P(E|F)$ and $P(F|E)$.

Why it's wrong: the event after the bar is the condition, so its probability is the denominator. Using $P(E)$ under $P(E|F)$ gives $0.2/0.6$ instead of $0.2/0.3$. Correct approach: $P(E|F)=\frac{P(E\cap F)}{P(F)}$, $P(F|E)=\frac{P(E\cap F)}{P(E)}$.

Mistake 2: Expecting $P(E|F)=P(F|E)$.

Why it's wrong: conditional probability is not symmetric; here they are $\frac23$ and $\frac13$. Correct approach: compute each with its own denominator.