Conditional Probability: Numerical Problems Set (7)

Question

A box contains 3 white and 5 black balls. Two balls are drawn one after the other without replacement. Find the probability that the second ball is white given that the first ball is black.

Solution — Step by Step

Why This Works

Conditional probability is just probability computed in the smaller sample space where the condition is true. Once we know the first ball is black, we recompute everything as if the bag now has 7 balls (3 white + 4 black). Whatever event we ask about is evaluated against this updated bag.

Formally, $P(B|A) = P(A \cap B) / P(A)$. We can verify: $P(A \cap B) = (5/8)(3/7) = 15/56$. $P(A) = 5/8$. Ratio: $(15/56)/(5/8) = (15/56)(8/5) = 3/7$. Same answer.

Alternative Method

Use the formal definition:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{(5/8)(3/7)}{5/8} = \frac{3}{7}$

This longer route is useful when the conditional reduction isn’t intuitive.

Common Mistake

Computing $P(A \cap B) = P(A) \times P(B)$ as if the events were independent. They’re not — the second draw depends on the first. Independence requires $P(B|A) = P(B)$, which means the condition shouldn’t change the probability. With replacement, draws are independent; without replacement, they’re not.