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 without 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

we have 8+7+10 = 25

if the first is coke then 8/25

if the second is sun kist 7 / 24

if the thirs is root beer 10 / 23

so the probability is 8/25 * 7/24 * 10/23 = 0.0406