Question

Use the following information to answer the next two exercises. 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, with replacement.

Let G1 = first card is green

Let G2 = second card is green

a. Draw a tree diagram of the situation.

b. Find P(G1 AND G2).

c. Find P(at least one green).

d. Find P(G2|G1).

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

Short answer

a. Tree diagram is below:

b. $\frac{25}{64}$

c. $\frac{55}{64}$

d. $\frac{5}{8}$

e. Yes.

Step-by-step solution

Given information

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, with replacement.

Let G1 = first card is green

Let G2 = second card is green

Part (a) Step 1: Explanation 

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 $P\left(G_2\right)=\frac{5}{8}$

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

The tree diagram of the situation is well-explained below

Part (b) Step 1: Explanation 

$P(G_{1}AND{G}_2)=\frac{5}{8}\times \frac{5}{8}=\frac{25}{64}$

Part (c) Step 1: Explanation 

Probability of at least one green

$$\begin{gathered} P(atleastonegreen)=P\left(GG\right)+P\left(GY\right)+P\left(YG\right) \\ P(atleastonegreen)=\frac{5}{8}\times \frac{5}{8}+\frac{5}{8}\times \frac{3}{8}+\frac{3}{8}\times \frac{5}{8} \\ P(atleastonegreen)=\frac{25}{64}+\frac{15}{64}+\frac{15}{64} \\ P(atleastonegreen)=\frac{55}{64} \end{gathered}$$

Part (d) Step 1: Explanation 

$$\begin{gathered} P\left(\overline{)G_1}{G}_2\right)=\frac{P(G_{1}AND{G}_2)}{P\left(G_2\right)} \\ P\left(\overline{)G_1}{G}_2\right)=\frac{{\displaystyle \frac{25}{64}}}{{\displaystyle \frac{5}{8}}}=\frac{5}{8} \end{gathered}$$

Part (e) Step 1: Explanation 

Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.