Question

A bottled water company runs a promotion in which 1 out of every 5 bottles has the word "Winner" printed under the

cap. Winners receive a free bottle of water. A store owner notices that in the last 8 bottles of water purchased, 3 have

been winners. What is the probability of getting 3 Winner" caps on 8 bottles of water?

Perspectives

Answer 1

The answer is "0.15".

Explanation:

From the information provided, they can assume that Binomial distribution should proceed, after all, "success". Its bottle cap winner as well as the result is also independent of another trail, but trails are not fixed at all. Yeah, that's how it works. We will claim also that chances of success are P=0.20 so there are no trials n=8.

Let an X=winner probability distribution be on the bottle cap, but this X is a binomial function of P=0.20, but n = 8

That is
`X \sim \ B(0.20,8)`

There we must find the chance that from 8 trials, this is exactly 3 winners limit that is P(X=3) if

`X\sim B(n,P) \ then \ P(X=x) \binom{n}{x} P^(x)(1-P)^(n-x)`

Therefore,
`P(X=3)= \binom{8}{3} 0.20^(3)(1-0.20)^(8-3)`

`= \binom{8}{3} 0.20^(3) (1-0.20)^(5)\\\\= \binom{8}{3} 0.20^(3) (0.80)^(5)\\\\=0.14680064 \\\\ = 0.15`