# Manhattan GMAT Challenge Problem of the Week – 5 Oct 2010

by on October 5th, 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

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.

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• The correct answer is 'C'
2->(1,1)
3->2* (1,2)
5->2*((1,4)(2,3))
7->2*((3,4) (2,5) (1,6))
11->2*(5,6)
total = 15
two dice when rolled gives a total of 36 outcomes.

so 15/ 36
= 5/12.

Thanks
Hari

• The correct answer is D

Here is the explanation.

Following are the possibilities to get the prime no.

1 cube 2nd Cube
1 2,4,6
3 2,5,4
6 5

Same with intechanging the cube. So total no of possibilities are 14 to get the addition of no as prime no.

Total no of combination with two cube are 36

So probability is 14/36 = 7/18

• Here is the explanation.

1st Cube 2nd Cube
1 2, 4, 6
3 2, 5, 4
6 5

Same will be interchange between two cube, Hence total no of ways to get the prime no summation is 14 (7+7)

Now always with two total no of combinations are 36

So prob. = 14/36 = 7/18

• Is there any other way then listing the outcomes manually??

• I thought that if it landed on a and then b that they both had to be different and we were excluding two of the same number..

Unfortunately there is no algebraic "shortcut" for the number of rolls that give a prime number (since there is no pattern in the occurrence of prime numbers) so we would have to list them. And since there is no restriction that A doesn't equal B, we have to accept rolling "doubles".

BIG ASIDE - This is ONLY for you number-junkies out there!
There is a **fun** little pattern that emerges.

Sum of the Numbers:
2 - (1,1)
3 - (1,2) (2,1)
4 - (1,3) (3,1) (2,2)
5 - (1,4) (4,1) (2,3) (3,2)
6 - (1,5) (5,1) (2,4) (4,2) (3,3)
7 - (1,6) (6,1) (2,5) (5,2) (3,4) (4,3)
8 - (2,6) (6,2) (3,5) (5,3) (4,4)
9 - (3,6) (6,3) (4,5) (5,4)
10 - (4,6) (6,4) (5,5)
11 - (5,6) (6,5)
12 - (6,6)

Thankfully we didn't have to list out all of the possible values, but sometimes it can pay to recognize that the number of times each event happens (sum=2, sum=3) will build in a symmetrical manner. This happens with the difference of 2, sum of 4 or even 6 times the sum of 20 dice.

• Unfortunately there is no algebraic "shortcut" for the number of rolls that give a prime number (since there is no pattern in the occurrence of prime numbers) so we would have to list them. And since there is no restriction that A doesn't equal B, we have to accept rolling "doubles".