Question

(FIXED) .Three symmetrical and homogeneous dice are tossed simultaneously. Find the probability of event A: the total number of dots appears at 10, given event B is at least one dice has upper face 3 dots. (the result of lesson is p(A|B) = p(AB)/p(B) = (15/6^3)/(91/6^3))

Answer 1

P(A | B) = P(A and B) / P(B)

P(A and B) = Probability the total number of dots appears at 10 and at least one dice has upper face 3 dots

= Probability that total number of dots appears at 7 for 2 dices and probability that any of three dices have 3

Number of possible combinations of numbers = (1, 3, 6) , (2, 3, 5) , (3, 3, 4)

For (1, 3, 6), number of possible permutations in 3 dices = 3! = 6

For (2, 3, 5), number of possible permutations in 3 dices = 3! = 6

For (3, 3, 4), number of possible permutations in 3 dices (2 unique numbers) = ³P₂ = 3

Total possible combinations = 6 + 6 + 3 = 15

Probability that total number of dots appears at 7 for 2 dices and probability that any of three dices have 3= 15 / 6^3

Probability that at least one dice has upper face 3 dots = 1 - Probability that no dice has upper face 3 dots

= 1 - (5/6) * (5/6) * (5/6) = 1 - 125 / 216

= (216 - 125) / 216 = 91/6^3

P(A | B) = P(A and B) / P(B) = 15 / 6^3 / 91/6^3

= 15/91