# Thread: Another binomial question

1. ## Another binomial question

A company distributed a large number of lottery tickets to their customers. It is known that 20% of the tickets are winning tickets which can be exchanged for a prize. Each customer is entitled to only one lottery ticket.

(i) A group of 15 customers is randomly chosen. Show that the probability that at least 3 customers will each win a prize is 0.602.
Hence, find the probability that at most 5 customers will each win a prize given that at least 3 customers will each win a prize.

(ii) Find the probability that among three randomly chosen groups of 15 customers, there is one group with at least 3 customers that will each win a prize and two other groups with not more than 2 customers that will each win a prize.

How should I go about doing part (ii) of this question?

Heres my attempt at it:
First, I found the probability of:
1) P(X>=3) [at least 3 customers] and
2) P(X<=2) [not more than 2 customers]

P(X>=3)= 0.602
P(X<=2)= 0.3980

and then i got stuck.

THANK YOU!

2. ## Re: Another binomial question

Originally Posted by pumbaa213
A company distributed a large number of lottery tickets to their customers. It is known that 20% of the tickets are winning tickets which can be exchanged for a prize. Each customer is entitled to only one lottery ticket.

(i) A group of 15 customers is randomly chosen. Show that the probability that at least 3 customers will each win a prize is 0.602.
Hence, find the probability that at most 5 customers will each win a prize given that at least 3 customers will each win a prize.

(ii) Find the probability that among three randomly chosen groups of 15 customers, there is one group with at least 3 customers that will each win a prize and two other groups with not more than 2 customers that will each win a prize.

How should I go about doing part (ii) of this question?

Heres my attempt at it:
First, I found the probability of:
1) P(X>=3) [at least 3 customers] and
2) P(X<=2) [not more than 2 customers]

P(X>=3)= 0.602
P(X<=2)= 0.3980

and then i got stuck.

THANK YOU!
because they've selected the two different events of interest

(X>=3) and (X<=2) as complementary you can again view this as a binomial distribution.

In this case p=Pr[x>=3] =0.602, and (1-p)=Pr[X<3]=0.398

so Pr[1 (X>=3) and 2 (X<3)] = $\displaystyle C(3,1) p^1(1-p)^2$ where $\displaystyle p=0.602$ and $C(n,k)$ is n choose k