Challenge Problem 09/27/10
Question
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.
quickly listing the first roll, then the successful second rolls:
First roll = 1: second roll = 1, 2, 4, 6
My Question
When I approached this I did not include the possibility of (roll1) 1 + (roll2) 2 = prime number 2. I excluded it as the question stem states the cube "is rolled twice, first landing on a and then landing on b."
Am I incorrect in assuming that 'a' and 'b' are distinct sides, and thus cannot both be the side with value =1?
Any help is much appreciated