A bag contains 5 white marbles and 3 black marbles. A marble is drawn and not replaced. Then, a second marble is drawn. Find the probability of drawing two white marbles.

Bag contains 5 white marbles (W) and 2 black marbles (B).
Total marbles = $5 + 2 = 7$.

Probabilities on first draw:
P(W) = $\frac{5}{7}$, P(B) = $\frac{2}{7}$.

Marble is replaced, so probabilities on second draw are the same:
P(W) = $\frac{5}{7}$, P(B) = $\frac{2}{7}$.

The tree diagram branches are:

First draw

$$\quad \swarrow \frac{5}{7} \quad (W)$$ $$Start \searrow \frac{2}{7} \quad (B)$$

Second draw from first W:

$$\frac{5}{7} \quad (W) \qquad \frac{2}{7} \quad (B)$$

Second draw from first B:

$$\frac{5}{7} \quad (W) \qquad \frac{2}{7} \quad (B)$$

The four paths and their probabilities:
WW: $\frac{5}{7} \times \frac{5}{7} = \frac{25}{49}$
WB: $\frac{5}{7} \times \frac{2}{7} = \frac{10}{49}$
BW: $\frac{2}{7} \times \frac{5}{7} = \frac{10}{49}$
BB: $\frac{2}{7} \times \frac{2}{7} = \frac{4}{49}$

Formula: P(two whites) = P(first W and second W)
= P(first W) $\times$ P(second W)

Substitute: P(first W) = $\frac{5}{7}$, P(second W) = $\frac{5}{7}$ (replacement)

Simplify: $\frac{5}{7} \times \frac{5}{7} = \frac{25}{49}$

$\dfrac{25}{49}$
\dfrac{25}{49}

Formula: P(one W and one B) = P(W then B) + P(B then W)

First, P(W then B) = P(first W) $\times$ P(second B)
Substitute: $\frac{5}{7} \times \frac{2}{7}$
Simplify: $\frac{10}{49}$

Next, P(B then W) = P(first B) $\times$ P(second W)
Substitute: $\frac{2}{7} \times \frac{5}{7}$
Simplify: $\frac{10}{49}$

Total: $\frac{10}{49} + \frac{10}{49} = \frac{20}{49}$

$\dfrac{20}{49}$
\dfrac{20}{49}