There are 9 coins. 1,1,2,2,2,3,3,3,3. What is the probability of choosing three coins that add up to 6?

Question

There are 9 coins. 1,1,2,2,2,3,3,3,3. What is the probability of choosing three coins that add up to 6?

Answer 1

Hi Omar,

 

I would break this problem into cases. There are only two (unordered) triplets that sum to 6 using 1,1,2,2,2,3,3,3,3. Here's why:

 

Case 1 - one of the three numbers is 1. Then the remaining two numbers must sum to 5, which can only occur as 2+3. This makes the triplet (1,2,3).

Case 2 - none of the three numbers is 1. Then 3 cannot be one of the three numbers because 3+3 = 6 (too large) and 3+2 = 5 (off by 1). This makes the triplet (2,2,2).

 

Now to compute probabilities. Think of each of the 9 numbers as distinct, so there are 9C3 = 84 ways to select 3 numbers. Now think of how many ways there are to select one each of 1, 2, and 3. This happens in 2C1 × 3C1 × 4C1 = 24 ways. Finally, how many ways are there to select all three 2's? 3C3 = 1 way.

 

P(sum of 6) = (24+1)/84