A.1 Urn A contains 3 red and 5 black balls, whereas urn B contains 6 reds and 4

Question

A.1 Urn A contains 3 red and 5 black balls, whereas urn B contains 6 reds and 4 white balls. a) f a bal 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 c) If a ball is randomly selected from each urn, what is the probability that the selected two d) If two balls will be randomly selected from one of the urns without replacement, what is e) Suppose that we win S3 for each red ball selected, lose S1 for each black ball selected selected bal is from urn B? balls are different color? the probability that the selected two balls are in red color? 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

a) Here, we are given that:

P( urn A) = P( urn B) = 0.5,

P( red | urn A) = 3/8 = 0.375
P( red | urn B) = 6/10 = 0.6

Using law of total probability, we get:

P( red ) = 0.5*( 0.375 + 0.6 ) = 0.4875

Using bayes theorem, we get:

P( urn A) = P( red | Urn A) P( Urn A) / P( red ) = 0.5*0.375 / 0.4875 = 0.3846

Therefore 0.3846 is the required probability here.

b) Now we know that: P( red ) = 0.4875, therefore P( black ) = 1 - P( red ) = 1 - 0.4875 = 0.5125

Using bayes theorem, we get:

P( Urn B | white ) = P( white | Urn B) P( Urn B) / P( white ) = 0.4*0.5 / 0.5125 = 0.3902

Therefore 0.3902 is the required probability here.

c) Probability that the selected two balls are different colour

= Probability that black is drawn from first urn and red from second urn + Probability that red is drawn from first urn and black from second urn

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

= 0.525

Therefore 0.5250 is the required probability here.

d) Probability that both balls selected are red is computed as:

= Probability that two balls are drawn from first urn * Probability that 2 red balls are drawn given that they are drawn from first urn + Probability that two balls are drawn from second urn * Probability that 2 red balls are drawn given that they are drawn from second urn

= 0.5*(3/8)*(2/7) + 0.5*(6/10)*(5/9)

= 0.2202

Therefore 0.2202 is the required probability here.

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