The sides of each of two plastic cubes are numbered 1 through 6, and

Question

The sides of each of two plastic cubes are numbered 1 through 6, and each number is equally likely to appear face up after either cube is rolled. What is the probability that, after the two cubes are rolled, the sum of the two numbers appearing face up will be greater than 9?

(A) \( \frac{1}{6} \)

(B) \( \frac{1}{5} \)

(C) \( \frac{1}{4} \)

(D) \( \frac{5}{18} \)

(E) \( \frac{1}{3} \)

(A)   \frac{1}{6}

(B)   \frac{1}{5}

(C)   \frac{1}{4}

(D)   \frac{5}{18}

(E)   \frac{1}{3}

Bunuel · Answer

When rolling two dice, there are 6 * 6 = 36 possible outcomes. To get a sum greater than 9, we look at the following combinations:

(6, 4)
(4, 6)

(6, 5)
(5, 6)

(6, 6)

(5, 5)

So the probability is 6/36 = 1/6.

Answer: A.

APram · Answer

I did it the following way:
If cube 1 is 1,2,3 then no way the sum will be greater than 9.

If cube 1 is 4, for sum to be greater than 9, cube 2 has only 1 option ie 6
So probability =  1/6 x 1/6

If cube 1 is 5, for sum to be greater than 9, cube 2 has 2 option ie 5, 6
So probability =  1/6 x 2/6

If cube 1 is 6, for sum to be greater than 9, cube 2 has 3 option ie 4, 5, 6
So probability =  1/6 x 3/6

Total =  1/36 + 1/18 + 1/12 = 1/6

DanTheGMATMan · Answer

Best to list out the explicit possibilities, how you sum to 10, 11, or 12, and put that number over the total
 https://www.youtube.com/watch?v=kBsFCEzcKWY

GMATinsight · Answer

Sum greater than 
10 = (4, 6) (5, 5) (6, 4) = 3 cases
11 = (5, 6) (6, 5) = 2 cases
12 = (6, 6) = 1 case

Favorable cases = 3+2+1 = 6

Total Outcomes = 6*6 = 36

Probability = 6/36 = 1/6

Answer: Option A

Related Video:

https://www.youtube.com/watch?v=29Rl7wUUOQ8

MartyMurray · Answer

The sides of each of two plastic cubes are numbered 1 through 6, and each number is equally likely to appear face up after either cube is rolled. What is the probability that, after the two cubes are rolled, the sum of the two numbers appearing face up will be greater than 9?

(A) \( \frac{1}{6} \)

(B) \( \frac{1}{5} \)

(C) \( \frac{1}{4} \)

(D) \( \frac{5}{18} \)

(E) \( \frac{1}{3} \)

To calculate the probability, we need to find the number of possible favorable outcomes and the total number of possible outcomes.

Find the total:

6 × 6 = 36

Find the favorable:

To find the favorable, we basically have to use a brute force method, finding pairs that work by going through the possibilities in an organized way.

4 6

5 5
5 6

6 4
6 5
6 6

Favorable = 6

\frac{Favorable}{Total} = \frac{6}{36} = \frac{1}{6}

(A)   \frac{1}{6}

(B)   \frac{1}{5}

(C)   \frac{1}{4}

(D)   \frac{5}{18}

(E)   \frac{1}{3}

Correct answer:   A

joyousn · Answer

Why isnt 6, 6 and 5,5 counted twice if there are two separate instances where you can get that number? sorta like 6 , 4 and 4,6

Bunuel · Answer

Pairs like (6,4) and (4,6) are different because the first die can be 6 and the second 4, or the first die can be 4 and the second 6. Those are two distinct outcomes among the 36 equally likely ones. However, for (6,6), both dice must show 6, which means the first die is 6 and the second die is 6. There is no second option or second case here, so it counts only once.

MartyMurray · Answer

If you think about it, there aren't really two separate instances for each.

We have two cubes that are rolled separately.

The only way to get 6, 6 is for the first cube to show 6 and then the second cube to show 6. There's no second way to get 6, 6. 5, 5 works the same way.

So, there's just one way for each of those outcomes to happen.

The sides of each of two plastic cubes are numbered 1 through 6, and

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The sides of each of two plastic cubes are numbered 1 through 6, and

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

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My Rewards