Question

An urn contains 3 white balls and 7 red balls. A second urn contains 8 white balls and 2 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? (Round your answers to three decimal places.)
(a) the probability that the urn selected was the first one

(b) the probability that the urn selected was the second one

Answer 1

Answer:

A) 0.36

B) 0.64

Step-by-step explanation:

Given data:

1st urn consist of 3 white, and seven red ball

2nd urn consist of  8 white and 2 red ball

P(A) = 0.8 given

therefore P(B) = 1 -0.8 = 0.2

P(W/A)  white ball from 1st urn $=  \frac{3}{10} \times  \frac{3}{10}$

P(W/B)  white ball from second urn $=  \frac{8}{10} \times  \frac{8}{10}$

a)$P(B/W)  = \frac{ P(A) \times  P(W/A)}{P(A) \times  P(W/A) + P(B) \times  P(W/B)}$

                $= \frac{0.8\times \frac{9}{100}}{0.8\times \frac{9}{100} + 0.2\times \frac{64}{100}}$

                  = 0.36

$b) P(B/W) =  \frac{ P(B) \times  P(W/B)}{P(B) \times  P(W/B) + P(A) \times  P(W/A)}$

                 $= \frac{0.2\times \frac{64}{100}}{0.2\times \frac{64}{100} + 0.8\times \frac{9}{100}}$

                  = 0.64

Answer 2

The probability that the urn selected was the first one is 0.36. and the probability that the urn selected was the second one is 0.64.

What is probability?

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

Given

A first urn contains 3 white balls and 7 red balls.

A second urn contains 8 white balls and 2 red balls.

P(A) = 0.8

Therefore

P(B) = 1 - 0.8 = 0.2

The white ball from the first urn will be

$\rm P(W/A) = \dfrac{3}{10}*\dfrac{3}{10}\\\\P(W/A) = \dfrac{9}{100}$

The white ball from the second urn will be

$\rm P(W/B) = \dfrac{8}{10}*\dfrac{8}{10}\\\\P(W/B) = \dfrac{64}{100}$

a)  The probability that the urn selected was the first one will be

$\rm P(B/W) = \dfrac{P(A)*P(W/A)}{P(A)*P(W/A)+P(B)*P(W/B)}\\\\P(B/W) = \dfrac{0.8*\frac{9}{100}}{0.8*\frac{9}{100} + 0.2 * \frac{64}{100}}\\\\P(B/W) = 0.36$

b)  The probability that the urn selected was the second one will be

$\rm P(B/W) = \dfrac{P(B)*P(W/B)}{P(A)*P(W/A)+P(B)*P(W/B)}\\\\P(B/W) = \dfrac{0.2*\frac{64}{100}}{0.8*\frac{9}{100} + 0.2 * \frac{64}{100}}\\\\P(B/W) = 0.64$

Thus, the probability that the urn selected was the first one is 0.36. and the probability that the urn selected was the second one is 0.64.

More about the probability link is given below.

https://brainly.com/question/795909