Confusion on simple probability questionSuppose you take a ball at random without knowing which urn you took it from. Here are two kinds of urns, one with three red balls and one blue, and one with three blue balls and one red. Both urns are equally likely. The ball turns out to be blue. What is the probability of getting two blue balls without knowing which bin they came from?Answer: 5/16I'm confused on how this answer is derived. My reasoning was that there are 8 balls and 4 are blue. The chance of drawing a blue ball the first time is 1 2 and the second time is still 1 2 (assuming replacement). 1 2 ∗ 1 2 = 1 4

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Confusion on simple probability questionSuppose you take a ball at random without knowing which urn you took it from. Here are two kinds of urns, one with three red balls and one blue, and one with three blue balls and one red. Both urns are equally likely. The ball turns out to be blue. What is the probability of getting two blue balls without knowing which bin they came from?Answer: 5/16I&#039;m confused on how this answer is derived. My reasoning was that there are 8 balls and 4 are blue. The chance of drawing a blue ball the first time is       1    2   and the second time is still       1    2   (assuming replacement).       1    2    ∗      1    2    =      1    4

Kaitlyn Levine · Accepted Answer

Let A1 denote the event &quot;first urn selected&quot;. Let A2 denote the event &quot;second urn selected&quot;. Let B denote the event &quot;both balls are blue&quot;.P(B|A1) = probability that both balls are blue given that the first urn was selected = (1/4)(1/4) = 1/16.P(B|A2) = probability that both balls are blue given that the second urn was selected = (3/4)(3/4) = 9/16.P(A1) = probability of selecting the first urn (note: unconditional probability, not the probability given you already know that the two balls are blue) = .5, this is given (it states the urns are equally likely to be chosen)P(A2) = probability of selecting the second urn (note: unconditional probability, not the probability given you already know that the two balls are blue) = .5, this is given (it states the urns are equally likely to be chosen)P(B) = probability of both balls being blue= P(B|A1)P(A1) + P(B|A2)P(A2) = (1/16)(1/2) + (9/16)(1/2) = 10/32 = 5/16Now you might ask, Given 2 blue balls were chosen, what is the probability the urn selected was the first one (A1) or the probability the urn selected was the second one (A2)?P(A1|B) = P(B|A1)P(A1)/P(B) = (1/16)(1/2)/(5/16) = 1/10 or about 10%P(A2|B) = 1 - P(A1|B) = 1 - 1/10 = 9/10 or about 90%The conclusion would be: Given that I picked 2 blue balls, there is about a 90% chance it was from A2, whereas only about a 10% chance it was from A1.. This make sense because A2 contains 3 blue balls compared to A1 which contains only 1.Cheers

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