Chapter 13.5, Problem 16E

To determine

To calculate: The probability that chip is from first bag provided that it is a blue chip.

Answer to Problem 16E

The probability that chip is from first bag provided that it is a blue chipis $\frac{2}{5}$ .

Explanation of Solution

Given information:

First bag contains 4 red chips and 4 blue chips. Second bag contains 2 red chips and 6 blue chips.

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 thatfirst bag contains 4 red chips and 4 blue chips. Second bag contains 2 red chips and 6 blue chips.

Total number of chips are 16. There are 6 red chips and 10 blue chips.

The probability that chip is from first bag provided that it is a blue chip.

Let A be an event that chip is drawn from first bag and B be an event that blue chip is drawn.

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

Probability that blue chip is drawn,

  $P\left(B\right)=\frac{10}{16}$

Probability that chip is from first bag and it is a blue chip,

  $P\left(A\text{ and B}\right)=\frac{4}{16}$

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 chip is from first bag provided that it is a blue chip.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{4}{16}}{\frac{10}{16}}\\ =\frac{4}{10}\\ =\frac{2}{5}\end{array}$

Thus, the probability that chip is from first bag provided that it is a blue chipis $\frac{2}{5}$ .