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Two fair six-sided dice are rolled. What is the probability that the s 30 Nov 2022, 13:36
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55% (hard)

Question Stats:

71% (02:18) correct 29% (02:39) wrong based on 249 sessions

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Two fair six-sided dice are rolled. What is the probability that the sum is either a perfect square or a perfect cube?

(A)\(\frac{5}{12}\)

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

(C)\(\frac{7}{18}\)

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

(E)\(\frac{1}{2}\)
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D
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Two fair six-sided dice are rolled. What is the probability that the s 30 Nov 2022, 14:28
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cat2010
Two fair six-sided dice are rolled. What is the probability that the sum is either a perfect square or a perfect cube?

(A)\(\frac{5}{12}\)

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

(C)\(\frac{7}{18}\)

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

(E)\(\frac{1}{2}\)
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Perfect square

  • 4 = (1,3), (2,2), (3,1)
  • 9 = (3,6), (4,5), (5,4), (6,3)

Perfect cube

  • 8 = (2,6), (3,5), (4,4), (5,3), (6,2)

Favorable Pairs = 12

Total Pairs = 36

Required Probability = 12/36 = 1/3

Option D
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Two fair six-sided dice are rolled. What is the probability that the s 15 May 2024, 13:05
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Why wasn't one considered as a perfect square/cube
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Two fair six-sided dice are rolled. What is the probability that the s 15 May 2024, 13:08
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hrishi84
Two fair six-sided dice are rolled. What is the probability that the sum is either a perfect square or a perfect cube?

(A)\(\frac{5}{12}\)

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

(C)\(\frac{7}{18}\)

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

(E)\(\frac{1}{2}\)­

Why wasn't one considered as a perfect square/cube
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1 is both a perfect square and a perfect cube; however, the sum of the values of two dice rolls cannot be 1.­
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Two fair six-sided dice are rolled. What is the probability that the s 13 Sep 2024, 05:32
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Given that 2 six-sided fair dice are rolled and we need to find what is the probability that the sum is either a perfect square or a perfect cube .

As we are rolling 2 dice => Number of cases = 6*6 = 36

Sum of numbers in the two dice will be in the range (1+1) = 2 and (6+6) = 12
=> Possible perfect squares between 2 and 12 are 4, 9
=> Possible perfect cubes between 2 and 12 are 8

Cases for sum to be perfect square:
  • Sum = 4 -> (1,3), (2,2), (3,1) -> 3 cases
  • Sum = 9 -> (3,6), (4,5) (5,4), (6,3) -> 4 cases

Cases for sum to be perfect cube:
  • Sum = 8 -> (2,6), (3,5), (4,4), (5,3), (6,2) -> 5 cases

=> Total cases = 3 + 4 + 5 = 12

=> P(Sum=Perfect Square or Cube) = \(\frac{12}{36}\) = \(\frac{1}{3}\)

So, Answer will be D
Hope it helps!

Playlist on Solved Problems on Probability here

Watch the following video to MASTER Dice Rolling Probability Problems

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