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IF 4 dice are thrown, what is the probability of getting

i.exactly 3sixes
ii.exactly2sixes
iii.no sixes.​

1 Answer

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When rolling a fair six-sided die, the probability of rolling a six is 1/6. Assuming that the dice are fair and independent, we can use the rules of probability to find the probability of rolling certain outcomes when 4 dice are thrown:

i. To find the probability of getting exactly 3 sixes when 4 dice are thrown, we can use the binomial probability formula:

P(exactly k successes in n trials) = (n choose k) * p^k * (1-p)^(n-k)

where n is the number of trials, k is the number of successful outcomes, p is the probability of success on each trial, and (n choose k) is the binomial coefficient.

For this problem, n = 4, k = 3, and p = 1/6.

P(exactly 3 sixes) = (4 choose 3) * (1/6)^3 * (5/6)^1

= 4 * (1/216) * (5/6)

= 5/54

Therefore, the probability of getting exactly 3 sixes when 4 dice are thrown is 5/54.

ii. To find the probability of getting exactly 2 sixes when 4 dice are thrown, we can use the same formula with k = 2:

P(exactly 2 sixes) = (4 choose 2) * (1/6)^2 * (5/6)^2

= 6 * (1/36) * (25/36)

= 25/72

Therefore, the probability of getting exactly 2 sixes when 4 dice are thrown is 25/72.

iii. To find the probability of getting no sixes when 4 dice are thrown, we can use the complement rule:

P(no sixes) = 1 - P(at least one six)

To find the probability of getting at least one six, we can use the complement of getting no sixes:

P(at least one six) = 1 - P(no sixes)

For each die, the probability of not rolling a six is 5/6. Therefore, the probability of getting no sixes on 4 dice is:

P(no sixes) = (5/6)^4

= 625/1296

Therefore, the probability of getting no sixes when 4 dice are thrown is 625/1296, and the probability of getting at least one six is 1 - 625/1296 = 671/1296.

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