Question

Suppose there are 3 bags, each containing 3 colored balls. The first two bags contain a red, green, and blue ball. The third bag contains 3 red balls. You choose a bag at random. You then pull out a ball and replace it back into the bag. You then pull out another ball and determine that both of the balls you chose were red. What is the probability you chose the bag with the 3 red balls?

Answer 1

Answer:

The probability you choose the bag with the 3 balls 81%.

Step-by-step explanation:

Given : Suppose there are 3 bags, each containing 3 colored balls. The first two bags contain a red, green, and blue ball. The third bag contains 3 red balls. You choose a bag at random. You then pull out a ball and replace it back into the bag. You then pull out another ball and determine that both of the balls you chose were red.

To find : What is the probability you chose the bag with the 3 red balls?

Solution :

The probability of three bags is

$P_1=P_2=P_3=\frac{1}{3}$

Let the event of red balls be E.

The probability of 2 red balls from first bag is $P(P_1/E)=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

The probability of 2 red balls from second bag is $P(P_2/E)=\frac{1}{9}$

The probability of 2 red balls from third bag is $P(P_3/E)=1$

The bag with the 3 red balls is

$P(E/P_3)=\frac{P(P_3/E)}{P(E)}$

$P(E/P_3)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{1}{9}+\frac{1}{3}\times \frac{1}{9}+\frac{1}{3}\times1}$

$P(E/P_3)=0.818$

The probability you choose the bag with the 3 balls 81%.