Question

Three digits are randomly selected without replacement from two 2s, three 6s and one 9. What is the probability that the sum of the three digits is a multiple of 2? Express your answer as a common fraction.

Answer 1

The probability that the sum of the three randomly selected digits is a multiple of 2 is 17/30.

Step-by-step explanation:

We can solve this problem by considering the different cases of even and odd numbers. There are a total of 6 digits: two 2s, three 6s, and one 9. Out of these, there are 4 even digits (the two 2s and two of the 6s) and 2 odd digits (one 6 and the 9).

To find the probability of selecting 3 digits that sum to a multiple of 2, we can consider three cases:

1. Selecting all even digits: There are 4 even digits to choose from, so the probability is 4/6.
2. Selecting 2 even digits and 1 odd digit: There are 4 even digits and 2 odd digits to choose from. We can choose 2 even digits in C(4,2) ways and 1 odd digit in C(2,1) ways. The probability is (C(4,2) * C(2,1)) / C(6,3).
3. Selecting all odd digits: There are 2 odd digits to choose from, so the probability is 2/6.

Adding up the probabilities from the three cases, the probability that the sum of the three digits is a multiple of 2 is:

(4/6) + ((C(4,2) * C(2,1)) / C(6,3)) + (2/6) = 4/6 + 12/20 + 2/6 = 34/60 = 17/30.