Probability is Almost One Third
Select two distinct positive integers (without order) between 1 and 29 inclusive. If the probability that their sum is divisible by 3 can be represented in the form b a where a and b are coprime positive integers, compute a + b .
The answer is 271.
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2 solutions
I used modular arithmetic, but pretty much the same solution
Did you notice that 2 0 3 6 8 is MORE than a third?
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"Almost" is referring to, in this context, that the probability is close to one-third, not necessarily that is a smidgen under one-third. (Where, as you pointed out, it is actually more than one-third.)
We have 3 types of nos: 3k, 3k+1 and 3k+2
From 1-29, there are 9 nos. of 3k types and 10 each of 3k+1 & 3k+2 type.
For sum of 2 nos. to be divisible by 3, either we should add 2 nos. of 3k types or 1 each of 3k+1 & 3k+2 type.
First can be done in 9 C 2 ways.
Second can be done in 1 0 C 1 x 1 0 C 1 ways.
Total number of ways to select 2 nos. out of 29 is 2 9 C 2
The desired probability is 4 0 6 3 6 + 1 0 0 = 2 0 3 6 8 .
So a+b = 271
The conditions for x + y to be divisible by 3 are;
(A) either x and y leave remainders of 1 and 2 or vice versa when divided by 3 or
(B) both x and y are divisible by 3 .
Between 1 and 29 inclusive, there are ten which leave a remainder of 1, ten which leave a remainder of 2 and nine which leave a remainder of 3 when divided by 3. So our probability is ( 2 2 9 ) ( 1 1 0 ) ( 1 1 0 ) + ( 2 9 ) = 2 0 3 6 8 Hence a + b = 6 8 + 2 0 3 = 2 7 1 .