Question

(MA-262 review) A fair six-sided die is rolled four times, and each result is recorded, in order. Determine (a) the probability that there are exactly two results (among the four) that are each a 3, and (b) the probability that the sum of the four results is 23. [Answers: 0.11574, 0.0030864.]

Answer 1

a) In the first case, we need to calculate the probability such that two of the throws contain the number 3 whose probability is 1/6 and the other two do not contain 3 whose probability is 5/6. And there are total $\binom{4}{2}$ such ways. Hence the answer becomes

$\binom{4}{2}$ * (1/6)² * (5/6)² = 25/216 = 0.11574

b) In the second case, we need the sumto be 23 which is possible only when three of the faces show 6 and one face shows up 5, and there are $\binom{4}{1}$ such ways. Hence the answer becomes

$\binom{4}{1}$​​​​​ * (1/6)⁴ = 4/1296 = 0.0030864