Urn A contains 3 red and 5 black balls, whereas urn B contains 6 reds and 4 whit

Question

Urn A contains 3 red and 5 black balls, whereas urn B contains 6 reds and 4 white balls.

a) If a ball is randomly selected and found to be red, what is the probability that the selected ball is from urn A?

b) If a ball is randomly selected and found to be white, what is the probability that the selected ball is from urn B?

c) If a ball is randomly selected from each urn, what is the probability that the selected two balls are different color?

d) If two balls will be randomly selected from one of the urns without replacement, what is the probability that the selected two balls are in red color?

e) Suppose that we win $3 for each red ball selected, lose $1 for each black ball selected and lose $2 for each white ball selected. If a ball is randomly selected from each urn, what is the probability that we will win the money?

Explanation / Answer

Let A and B be the event that the ball is selected from urn A and B respectively. Let R, W and C be the event that the ball selected is of colour red, white and black respectively.

P(A) = P(B) = (1/2) , as they are selected at random.

a)

P(R | A)  = 3 / (3+5) = 3/8

P(R | B) = 6 / (6+4) = 6 / 10

Probability that the selected ball is Red, P(R) = P(A) P(R | A) + P(B) P(R | B)

= (1/2) * (3/8) + (1/2) * (6/10) = 0.4875

Probability that the selected red ball is from urn A = P(A | R)

= P(R | A) P(A) / P(R) = (3/8) * (1/2) / 0.4875 = 0.3846

b)

P(W | A)  = 0

P(W | B) = 4 / (6+4) = 4 / 10

Probability that the selected ball is white, P(W) = P(A) P(W | A) + P(B) P(W | B)

= (1/2) * 0 + (1/2) * (4/10) = 0.2

Probability that the selected white ball is from urn B = P(B | W)

= P(W | B) P(B) / P(W) = (4/10) * (1/2) / 0.2 = 1

c)

The selected two balls are different color have the below combinations,

- Red ball from urn A and white ball from urn B

- Black ball from urn A and white ball from urn B

- Black ball from urn A and red ball from urn B

The probability that the selected two balls are different color = P(R|A) * P(W|B) + P(C|A) * P(W|B) + P(C|A) * P(R|B)

= (3/8) * (4/10) + (5/8) * (4/10) + (5/8) * (6/10) = 0.775

d)

If urn A is selected, probability that the selected two balls are in red color = (3/8) * (2/7) = 0.1071

If urn B is selected, probability that the selected two balls are in red color = (6/10) * (5/9) = 0.3333

So, probability that the selected two balls are in red color = (1/2) * 0.1071 + (1/2) * 0.3333 = 0.2202

e)

We will always win when we draw a red ball from urn A or urn B or both. So we will lose, if no red balls are drawn from both urns.

Probability that we will win the money = 1 - Probability of lose = 1 - Probability that no red balls are drawn from both urns = 1 - (5/8) * (4/10) = 0.75

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