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10 Apr 2026, 03:26
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C
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Difficulty:
55%
(hard)
Question Stats:
56% (02:06) correct
44%
(02:30)
wrong
based on 43
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Penelope rolls a fair six-sided die twice. What is the probability that the product of the two results is a multiple of 6?
(A)\( \frac{11}{36}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{25}{36}\)
(A)\( \frac{11}{36}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{25}{36}\)
10 Apr 2026, 07:30
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Penelope rolls a fair six-sided die twice. What is the probability that the product of the two results is a multiple of 6?
Calculate the total number of possible outcomes:
6 × 6 = 36
Calculate the number of favorable outcomes:
Have to use brute force because the number of favorable outcomes depends on what's rolled first.
1 + 2 + 3 + 2 + 1 + 6 = 15
15/36 = 5/12
(A)\( \frac{11}{36}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{25}{36}\)
Correct answer: C
Calculate the total number of possible outcomes:
6 × 6 = 36
Calculate the number of favorable outcomes:
Have to use brute force because the number of favorable outcomes depends on what's rolled first.
| First Roll | Second Roll | Total Outcomes |
| 1 | 6 | 1 |
| 2 | 3 or 6 | 2 |
| 3 | 2, 4, or 6 | 3 |
| 4 | 3 or 6 | 2 |
| 5 | 6 | 1 |
| 6 | 1, 2, 3, 4, 5, or 6 | 6 |
1 + 2 + 3 + 2 + 1 + 6 = 15
15/36 = 5/12
(A)\( \frac{11}{36}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{25}{36}\)
Correct answer: C
10 Apr 2026, 07:36
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Keeping 6 fixed we have from 1-5 each with 2 chances thus 5*2=10
We have 1 chance with 6*6.
We also have factors of 6 multiplying in:
2*3
3*4
Each accounting for 2 chances thus 4 in total.
10+1+4 = 15
15/36 = 5/12
Answer: Option C
We have 1 chance with 6*6.
We also have factors of 6 multiplying in:
2*3
3*4
Each accounting for 2 chances thus 4 in total.
10+1+4 = 15
15/36 = 5/12
Answer: Option C
kevincan
