Chapter 13, Problem 13STP

(a)

To determine

Two trials of the experiment are conducted.Find whether the events are independent or dependent.

Answer to Problem 13STP

Independent.

Explanation of Solution

Given:

A bag contains 3 red chips, 5 green chips, 2 yellow chips , 4 brown chips, and 6 purple chips.One chip is chosen at random , the color noted , and the chip returned to the bag.

Calculation:

The events here are drawing the chips from the bag.

The events are dependent if the occurrence of one event affects the outcome of the other event.

Since the chips are replaced in the bag, so , the events of drawing chips in two trials do not affect the outcome of each other.

So, the events are independent.

(b)

To determine

Find the probability that both the chips are purple.

Answer to Problem 13STP

0.09

Explanation of Solution

Given:

A bag contains 3 red chips, 5 green chips, 2 yellow chips , 4 brown chips, and 6 purple chips.One chip is chosen at random , the color noted , and the chip returned to the bag.

Formula Used:

  $P(A\text{ and }B)=P(A)\cdot P(B)$

Calculation:

The events of drawing the chips are independent.

Make a table of given information:

Color of Chips	Number of Chips
Red	3
Green	5
Yellow	2
Brown	4
Purple	6

Total number of chips = 3 + 5 + 2 + 4 + 6 = 20

Let

Event of drawing purple chip in 1^st trial = A

Event of drawing purple chip in 2^nd trial = B

  $P(A)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{6}{20}=\frac{3}{10}$

Since the drawn chip is replaced in the bag , so ,

  $P(B)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{6}{20}=\frac{3}{10}$

So, $P(A\text{ and }B)=P(A)\cdot P(B)=\frac{3}{10}\cdot \frac{3}{10}=\frac{9}{100}=0.09$

So, the probability that both the chips are purple is 0.09 .

(c)

To determine

Find the probability that first chip is green and second chip is brown.

Answer to Problem 13STP

0.05

Explanation of Solution

Given:

A bag contains 3 red chips, 5 green chips, 2 yellow chips , 4 brown chips, and 6 purple chips.One chip is chosen at random , the color noted , and the chip returned to the bag.

Formula Used:

  $P(A\text{ and }B)=P(A)\cdot P(B)$

Calculation:

The events of drawing the chips are independent.

Make a table of given information:

Color of Chips	Number of Chips
Red	3
Green	5
Yellow	2
Brown	4
Purple	6

Total number of chips = 3 + 5 + 2 + 4 + 6 = 20

Let

Event of drawing green chip in 1^st trial = A

Event of drawing brown chip in 2^nd trial = B

  $P(A)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{5}{20}=\frac{1}{4}$

Since the drawn chip is replaced in the bag , so ,

  $P(B)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{4}{20}=\frac{1}{5}$

So, $P(A\text{ and }B)=P(A)\cdot P(B)=\frac{1}{4}\cdot \frac{1}{5}=\frac{1}{20}=0.05$

So, the probability that both the chips are purple is 0.05 .