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A cube with its sides numbered 1 through 6 is rolled twice [#permalink]

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28 Apr 2012, 06:33

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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

I am not sure if the OA is correct. Thanks for help !

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

I am not sure if the OA is correct. Thanks for help !

Re: A cube with its sides numbered 1 through 6 is rolled twice [#permalink]

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13 Apr 2014, 02:13

Bunuel wrote:

qtrip wrote:

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

I am not sure if the OA is correct. Thanks for help !

The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?

Re: A cube with its sides numbered 1 through 6 is rolled twice [#permalink]

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13 Apr 2014, 02:21

NAL9 wrote:

Bunuel wrote:

qtrip wrote:

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

I am not sure if the OA is correct. Thanks for help !

The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?

Question says the roll landed on a and b, a and b can be different or same. The question does not give any idea about that, so assuming they are different is not correct in my opinion.

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

I am not sure if the OA is correct. Thanks for help !

The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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Re: A cube with its sides numbered 1 through 6 is rolled twice [#permalink]

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02 Sep 2015, 22:39

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Re: A cube with its sides numbered 1 through 6 is rolled twice [#permalink]

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30 Nov 2015, 19:10

I don't know if there is a short way to solve this, but I put down everything that can be obtained from the values of a and b: 1+1 = 2 -prime 1+2=3-prime 1+3=4-not 1+4=5-prime 1+5=5-not 1+6=prime

so when a is 1, the other must be 1,2, 4, or 6. the probability that a is 1 is 1/6. the probability that b is 1,2,4,6 is 4/6 or 2/3 the first scenario when a is 1, the overall probability that the a+b is prime is 1/6 * 2/3 or 1/9

when a is 2, in order for a+b to be prime, b must be: 1,3, or 5. the probability for a to be 2 is 1/6 and the probability of b to be 1,3, or 5 is 1/2 so overall probability for a+b to be prime when a is 2 is 1/6*1/2 or 1/12

when a is 3, b must be 2 or 4 and the overall probability is 1/6 * 1/3 = 1/18

when a is 4, b must be 1 or 3 -> overall probability 1/18

when a is 5, b must be 2 or 6 -> 1/18

when a is 6, b must be 1 or 5 -> 1/18

now, the sum of all probabilities:

1/9 + 1/12 + 4(1/18)

find LCM of 9, 12, and 18 -> 36 multiply first fraction by 4, second by 3, and the last by 2.

(4+3+8)/36 or 15/36. divide by 3 both numerator and denominator -> 5/12