Manhattan GMAT Challenge Problem of the Week – 5 Oct 2010

by Manhattan Prep, Oct 5, 2010

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A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0

(B) 1/12

(C) 5/12

(D) 7/18

(E) 4/9

Answer

This problem can be solved quickly by first listing the target numbers: primes that could be generated as the sum of two dice rolls.

Since the target numbers must be between 2 and 12 (inclusive), we have 2, 3, 5, 7, and 11.

Now go target by target, listing the possible rolls.

2: Roll 1, then 1. One way.

3: Roll 1, then 2.

Roll 2, then 1. Two more ways.

Realize that you have to separately count rolling a 1, then a 2 and rolling a 2, then a 1. Those are two separate ways to roll a 3.

5: 1, then 4.

2, then 3.

3, then 2.

4, then 1. Four ways.

7: 1, then 6.

2, then 5.

3, then 4.

4, then 3.

5, then 2.

6, then 1. Six ways.

11: 5, then 6.

6, then 5. Only two ways.

These ways sum up: 1+ 2 + 4 + 6 + 2 = 15. Divide by 36 (= 6 6) to get 15/36.

The correct answer is C.

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