A box contains two white and three black balls. Two balls…

A box contains two white and three black balls. Two balls are drawn one after another (without replacement). Find the probability that both balls are white.

Explanation of Method

The two draws are dependent: after a white ball is taken, fewer white balls remain. We use the multiplication rule $P(W_1W_2)=P(W_1)\,P(W_2\mid W_1)$, where the second factor is computed from the box as it is after the first draw. We check the result by counting pairs of balls.

Step 1: Events

$W_1$ = "the first ball is white", $W_2$ = "the second ball is white". We want $P(W_1W_2)$.

Step 2: First draw

$2$ white out of $5$: $P(W_1)=\frac25$.

Step 3: Second draw, given the first was white

Now the box has $1$ white and $3$ black balls: $P(W_2\mid W_1)=\frac14$.

Step 4: Multiplication rule

\[ \boxed{P(W_1W_2)=\frac25\cdot\frac14=\frac1{10}=0.1.} \]

Step 5: Check by counting

There are $\binom52=10$ equally likely pairs of balls, and only one of them consists of the two white balls: $\frac1{10}$. ✓

Result

	Answer
Formula	$P(W_1)P(W_2\mid W_1)=\frac25\cdot\frac14$
Probability	$0.1$