The same conditions as in the previous problem (two white…
The same conditions as in the previous problem (two white and three black balls, two balls drawn), but the first ball is returned to the box before the second draw.
Explanation of Method
Because the first ball is put back, the box is the same before each draw. So the two draws are independent: the result of the first does not change the chances of the second. The probability of two white balls is the product of the two (equal) probabilities.
Step 1: Each draw
Before each draw the box contains $2$ white and $3$ black balls: $P(W_1)=P(W_2)=\frac25$.
Step 2: Independence
Returning the ball restores the box, so $P(W_2\mid W_1)=P(W_2)=\frac25$: the events are independent.
Step 3: The answer
\[ \boxed{P(W_1W_2)=\frac25\cdot\frac25=\frac4{25}=0.16.} \]
Step 4: Check by counting ordered pairs
With replacement there are $5\cdot5=25$ equally likely ordered pairs, of which $2\cdot2=4$ are white–white: $\frac4{25}$. ✓
Step 5: Comparison with Problem 102
| Without replacement (102) | With replacement (103) | |
|---|---|---|
| $P(W_2\mid W_1)$ | $1/4$ | $2/5$ |
| $P(\text{both white})$ | $0.10$ | $0.16$ |
Returning the ball keeps the second white ball's chances higher, so two whites become more likely.
Result
| Answer | |
|---|---|
| Probability | $\left(\frac25\right)^2=0.16$ |