8 of one kind, 9 of another kind, and 10 on a third kind. What is the probability of picking 2 of the same kind on first 2 picks

Question

8 of one kind, 9 of another kind, and 10 on a third kind. What is the probability of picking 2 of the same kind on first 2 picks

Answer 1

There are 8 of the 1st kind, 9 of the 2nd kind and 10 of the 3rd kind.  There are 8 + 9 + 10 = 27 items total.

 

The way the problem is stated, I'm assuming that you are picking without replacement - so that when an item of a specific kind is picked, it is not put back - and not available to be used for the second pick.  So there are 27 items to choose from on the 1st pick and 26 items to choose from on the 2nd pick.

 

If we define A = the event of picking 2 of 1st kind

                  B = the event of picking 2 of 2nd kind

                  C = the event of picking 2 of 3rd kind

                  D = the event of picking 2 of the same kind on first 2 picks

 

then P(D) = P( A or B or C )

 

2 or more events are mutually exclusive if they cannot occur at the same time.  In this case, events A, B and C are mutually exclusive.  When events are mutually exclusive, the probability of the union: A or B or C is the sum of the individual probabilities.

 

That is, P( A or B or C ) = P(A) + P(B) + P(C)

 

P(A) = (8/27) * (7/26)

 

because once you pick 1 item of the 1st kind, there are only 7 items of the 1st kind remaining out of the 26 remaining items.  By similar reasoning, 

 

P(B) = (9/27) * (8/26)

P(C) = (10/27) * (9/26)

 

So P(D) = P(A) + P(B) + P(C) = (8*7 + 9*8 + 10*9)/(27*26) = 218/702 = 0.31 (rounded off) = 31%