91Ó°ÊÓ

Suppose that you have eight cards.

Five are green and three are yellow.

The cards are well shuffled.

Suppose that you randomly draw two cards, one at a time, without replacement.

Let G1 = first card is green

Let G2 = second card is green

The probability of first card is green is given by $P\left(G_1\right)=\frac{5}{8}$

The probability of second card is green is given by $P\left(G_2\right)=\frac{4}{7}$

The probability of first card is yellow is given by$P\left(Y_1\right)=\frac{3}{8}$

The probability of second card is yellow is given by $P\left(Y_2\right)=\frac{2}{7}$

The tree diagram of the situation is well-explained below

$$\begin{gathered} P(G_{1}AND{G}_2)=P\left(G_1\right).P\left(G_2\right) \\ P(G_{1}AND{G}_2)=\frac{5}{8}×\frac{4}{7} \\ P(G_{1}AND{G}_2)=\frac{20}{56} \end{gathered}$$

Probability of at least one green

$$\begin{gathered} P(atleastonegreen)=P\left(GG\right)+P\left(GY\right)+P\left(YG\right) \\ P(atleastonegreen)=\frac{5}{8}×\frac{4}{7}+\frac{5}{8}×\frac{2}{7}+\frac{3}{8}×\frac{4}{7} \\ P(atleastonegreen)=\frac{20}{56}+\frac{10}{56}+\frac{12}{56} \\ P(atleastonegreen)=\frac{42}{56} \end{gathered}$$

$$\begin{gathered} P\left(\overline{)G_1}{G}_2\right)=\frac{P(G_{1}AND{G}_2)}{P\left(G_2\right)} \\ P\left(\overline{)G_1}{G}_2\right)=\frac{{\displaystyle \frac{20}{56}}}{{\displaystyle \frac{4}{7}}}=\frac{5}{8} \end{gathered}$$

No, they are dependent because the first card is drawn in the bag before the second card is drawn; the composition of cards in the bag doesn't remain the same from draw one to draw two.