Question

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

$G1$= first card is green

$G2$= second card is green

a. Draw a tree diagram of the situation.

b. Find $P(G1ANDG2)$.

c. Find $P$(at least one green).

d. Find $P\left(G2\right|G1)$.

e. Are $G2$ and $G1$ independent events? Explain why or why not

Short answer

(a) The tree diagram is in the following steps

(b) $G1,G2$value is given, The value of $P\left(G1G2\right)$is equal to$\frac{20}{56}$

(c) The value of $P$(at least one green)$=\frac{50}{56}$

(d) $G1,G2$value is given, The value of $P\left(G_2\mid G_1\right)$is equal to $\frac{4}{7}$

(e)$G1$and$G2$are not independent because$G1$and$G2$values are not equal

Step-by-step solution

The tree diagram (part a)

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:

Find the value of P(G1G2) (part b)

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

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

To find the value of P(at least one green) (part c)

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)

$=\frac{3}{8}\times \frac{2}{7}$

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

$=\frac{50}{56}$

To find the value of P(G2|G1) (part d)

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

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

To find the G1 and G2 are independent or not (part e)

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}\times \frac{4}{7}$$+\frac{3}{8}\times \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.