Question

Three dice are thrown. The probability of getting a sum of numbers which is a perfect cube, is:

• A. \(\frac{1}{6}\)
• B. \(\frac{1}{8}\)
• C. \(\frac{5}{36}\)
• D. \(\frac{5}{36}\)

Answer 1

The Correct Option is D

To solve the problem of finding the probability that the sum of the numbers on three thrown dice is a perfect cube, first determine the possible sums that are perfect cubes. For three dice, each roll falls between 1 and 6, yielding possible sums between 3 (1+1+1) and 18 (6+6+6).

The perfect cubes within this range are 8 (2^3) and 27 (3^3). However, since 27 is not possible with three dice, the only viable sum is 8.

Now, calculate the number of outcomes that result in the sum being 8. Consider these combinations:

• (2,2,4)
• (2,3,3)
• (3,2,3)
• (4,2,2)
• (3,3,2)
• (3,2,3)

These offer 6 valid possibilities due to different permutations, each providing a sum of 8. Consequently, there are 10 favorable outcomes for getting a sum of 8.

Compute the probability: since each die roll is independent, the total outcomes from 3 dice are 6^3=216. Therefore, the probability is:

\(P(\text{sum is a cube}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{216}\)

Simplify the fraction:

\(= \frac{5}{36}\)

Thus, the probability of the sum being a perfect cube is \(\frac{5}{36}\).