Two distinct nonzero digits
were chosen randomly.
a. Given their sum is even, what is the probability
that one of the digits was 4?
b. Given one of the digits was 4, what is the probability
that their sum is even?
Overall, discounting the conditional probabilities asked for, there are 9*8=72 cases based on the first paragraph/sentence.
a. If the sum is even, the two digits are of the same parity. There are 5*4=20 where both are odd and 4*3=12 where both are even, for 32 cases where the total is even. Of these there are 6 cases where there is a 4: 4,2; 4,6; 4,8; 2,4; 6,4; 8,4. The answer to part a is 6/32 = 3/16.
b. Of the 72 cases in total, 8 cases have 4 as the first digit and 8 as the last; so 16 cases have a 4. We already know that there are 6 eventotalled ways that contain a 4, so the answer to part b is 6/16 = 3/8.

Posted by Charlie
on 20160513 14:46:05 