In a barrel there 8 cans of coke 7 cans of sun kist and 10 c

In a barrel there 8 cans of coke, 7 cans of sun kist, and 10 cans of root beer. If three cans are randomly selected with replacement, find the probability that the first selected is coke, the second selected is a sun kist, and the third selected is a root beer.

Solution

In a barrel there 8 cans of coke, 7 cans of sun kist, and 10 cans of root beer

three cans are randomly selected with replacement = first selected is coke * second selected is a sun kist * third selected is a root beer
= 8 C 1 / 25 C 1 * 7 C 1 / 25 C 1 * 10 C 1 / 25 C 1 = 8 * 7 * 10 / (25)^3 = 0.03584