Chapter 13.5, Problem 26E

To determine

To calculate: The probability that second drawn marble is green provided that first drawn marble was green.

Answer to Problem 26E

The probability that second drawn marble is green provided that first drawn marble was greenis $\frac{2}{7}$ .

Explanation of Solution

Given information:

There are 3 green marbles and 5 yellow marbles in a container. One marble is drawn and it not put in the container back then a second marble is drawn.

Formula used:

Probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information thatthere are 3 green marbles and 5 yellow marbles in a container. One marble is drawn and it not put in the container back then a second marble is drawn.

The probability that second drawn marble is green provided that first drawn marble was green.

Let A be an event that second drawn marble is green and B be an event that first marble drawn is green.

Recall that probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that first marble is green,

  $\begin{array}{c}P\left(B\right)=\frac{\text{number of green marbles}}{\text{total number of marbles}}\\ =\frac{3}{8}\end{array}$

Probability that second marble drawn is green and first drawn marble was green,

  $\begin{array}{c}P\left(A\text{ and B}\right)=\frac{3}{8}\cdot \frac{2}{7}\\ =\frac{3}{28}\end{array}$

Recall that given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability that second drawn marble is green provided that first drawn marble was green.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{3}{28}}{\frac{3}{8}}\\ =\frac{8}{28}\\ =\frac{2}{7}\end{array}$

Thus, the probability that second drawn marble is green provided that first drawn marble was greenis $\frac{2}{7}$ .