An urn contains 4 white balls and 6 red balls. A second urn contains 7 white balls and 3 red balls. An urn is selected, and the probability of selecting the first urn is 0.8. A ball is drawn from the selected urn and replaced. Then another ball is drawn and replaced from the same urn. If both balls are white, what are the following probabilities?

a. the probability that the urn selected was the first one
b. the probability that the urn selected was the second one

Answer:

a) 0.5664 = 56.64% probability that the urn selected was the first one.

b) 0.4336 = 43.36% probability that the urn selected was the second one

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

If both balls are white, what are the following probabilities?

This means that event A is both balls being white.

Probability of both balls being white:

0.8 probability of 4/10 = 0.4 squared(first urn, which each time the ball is drawn will have 4 white balls and 6 red)

1 - 0.8 = 0.2 probability of 7/10 = 0.07 squared(second urn). So

[tex]P(A) = 0.8*(0.4)^2 + 0.2*(0.7)^2 = 0.226[/tex]

a. the probability that the urn selected was the first one

Event B: First urn selected.

The intersection of events A and B is given by both balls being white and selected from the first urn, which has the following probability:

[tex]P(A \cap B) = 0.8*(0.4)^2 = 0.128[/tex]

The desired probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.128}{0.226} = 0.5664[/tex]

0.5664 = 56.64% probability that the urn selected was the first one

b. the probability that the urn selected was the second one

The intersection of events A and B is given by both balls being white and selected from the second urn, which has the following probability:

[tex]P(A \cap B) = 0.2*(0.7)^2 = 0.098[/tex]

The desired probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.098}{0.226} = 0.4336[/tex]

0.4336 = 43.36% probability that the urn selected was the second one

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