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24 Aug 2024, 10:31
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A fair six-sided die is rolled four times. What is the probability that at least one multiple of 3 is rolled?
A) 8/81
B) 16/81
C) 1/12
D) 65/81
E) 80/81
A) 8/81
B) 16/81
C) 1/12
D) 65/81
E) 80/81
24 Aug 2024, 10:37
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3 and 6 = multiples of 3.
4/6 or 2/3 chance a multiple of 3 is not rolled.
1 - (2/3)^4 = 65/81
4/6 or 2/3 chance a multiple of 3 is not rolled.
1 - (2/3)^4 = 65/81
21 Jul 2025, 07:28
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A fair six-sided die is rolled four times. What is the probability that at least one multiple of 3 is rolled?
Probability of getting at least one multiple of 3 = 1 - Probability of getting no multiples of 3
Out of the 6 numbers only two (3, 6) are multiples of 3
=> Probability of not getting a multiple of 3 in one roll = \(\frac{4}{6}\) = \(\frac{2}{3}\)
=> Probability of not getting a multiple of 3 in all 4 rolls = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) = \(\frac{16}{81}\)
=> Probability of getting at least one multiple of 3 = 1 - Probability of getting no multiples of 3 = 1 - \(\frac{16}{81}\) = \(\frac{81 - 16}{81}\) = \(\frac{65}{81}\)
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
Probability of getting at least one multiple of 3 = 1 - Probability of getting no multiples of 3
Out of the 6 numbers only two (3, 6) are multiples of 3
=> Probability of not getting a multiple of 3 in one roll = \(\frac{4}{6}\) = \(\frac{2}{3}\)
=> Probability of not getting a multiple of 3 in all 4 rolls = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) = \(\frac{16}{81}\)
=> Probability of getting at least one multiple of 3 = 1 - Probability of getting no multiples of 3 = 1 - \(\frac{16}{81}\) = \(\frac{81 - 16}{81}\) = \(\frac{65}{81}\)
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
