91Ó°ÊÓ

We have a total of eight cards. Five of them are yellow, and three are green.

The event $G1$is defined as when the first card is green.

$G2$refers to when the second card is also green.

The tree diagram is given by:

We need to figure out what the chances are that the second card will be green.

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

We need to figure out how likely it is that at least one card is green. That is the inverse of the situation where none of the cards are green.

$P$(at least one green)$=1-P$(no green cards)

$=1-P$(both cards are yellow)

$=\frac{3}{8}×\frac{2}{7}$

$=1-\frac{6}{56}$

$=\frac{50}{56}$

Given that the first card is green, we must calculate the likelihood that the second card will also be green.

$$\begin{gathered} P\left(G_2âˆ\pounds G_1\right)=\frac{P\left(G_1\mathrm{AND}{G}_2\right)}{P\left(G_1\right)} \\ =\frac{\frac{20}{56}}{\frac{5}{8}} \\ =\frac{4}{7} \end{gathered}$$

If $G1$and $G2$are self-contained, it follows that:

$P\left(G2\right|G1)$$=P\left(G2\right)$

Since the probability $P\left(G2\right|G1)$equals $\frac{7}{4}$and the probability $P\left(G2\right)equals$:

$P\left(G2\right)=$$P$(green and green) $+P$(yellow and green)

$=\frac{5}{8}×\frac{4}{7}$$+\frac{3}{8}×\frac{5}{7}$

$=\frac{35}{56}$

$\frac{35}{56}notequalto\frac{4}{7}$, as we know. As a result, we can conclude that$G1andG2$ are not mutually exclusive.