Homework - Part 1

I recently received this email:

Hi, Mr. Catlin,

Please if you can

help me with these questions. Thank you very much.

Question 1

An urn contains b black balls and r red balls. One of the balls is drawn at random. If it is a black ball, we put it back and add c additional black balls to the urn; when a red ball is drawn we put back the red ball. Then we draw a second ball. Compute the probability that the first ball drawn was black given that the second ball drawn was red.

Question 2

An ordinary, well shuffled deck is dealt to four people. Each of the players (A, B, C, D) receives 13 cards.

1. Find the probability that someone at the table has four Aces.
2. Find the probability that player A is void in at least one suit.

Now, I have been around the classroom for enough years that I recognize textbook questions when I see them. The writer is a student and wants me to do his/her homework. I wrote back and said that I would address these questions in a future article and asked the person for their name. No reply!

They are interesting questions, especially question 2 since it might be of interest to bridge players. I'll get to that next month.

For now I'll address question 1. This looks very similar to Polya's Urn Scheme for the notation is exactly the same in both cases. (See my October 2008 article in the archives.) It could be that the writer misstated it or it could be that the instructor changed Polya's scheme to challenge the class. In any case let me show you how to solve the writer's problem; I'll leave the algebra to you.

If A and B are events then P(A|B) is the conditional probability that A occurs given that B occurs. Letting P(A & B)represent the probability that both A and B occur we have the relationship

P(A & B) = P(A|B)P(B) (1)

We also have

P(A & B) = P(B|A)P(A) (2)

From (1) and (2) we can conclude that

P(A|B)P(B) = P(B|A)P(A) (3)

hence

P(B|A) = P(A|B)P(B) / P(A) (4)

Relation (4) is sometimes called Baye's theorem.

Now let me introduce the following events: B1 is the event that a black ball is drawn on the first draw, R1 is the event that a red ball is drawn on the first draw, and R2 is the event that a red ball is drawn on the second draw. From (4) we have that

P(B1|R2) = P(R2|B1)P(B1) / P(R2) (5)

Now the only slightly tricky part is calculating P(R2). Since either B1 or R1 occurs on the first draw we condition on these two events as follows

P(R2) = P(R2|B1)P(B1) + P(R2|R1)P(R1) (6)

Obviously

P(R2|B1) = r / (r + b + c)
P(R2|R1) = r / (r +b)
P(B1) = b / (r + b)
P(R1) = r / (r + b)

Substituting these into (6) and then into (5) and simplifying the algebraic expression we obtain

P(B1|R2) = b / (b +rh)

Where h is given by

h = 1 + c /(r + b)

I'll see you next month with question 2.

Don Catlin can be reached at 711cat@comcast.net