Statistics help

Question

If you buy a meal at Burger Bash, there is a 34% chance of winning a free hamburger. If 6 friends meet for lunch at Burger Bash, and they each buy a meal, what is the probability that exactly 2 of them will win a free burger?

Answer 1

Hi Emily,

This is similar to the problem you posted about the flights arriving on time; also a binomial case, also two outcomes: "success" (winning a burger) and "failure" (not). I'll set this up and leave the computations to you:

P(X=x)= C(n, x)*p^x*q^n-x

x=desired number of free burgers

n=total sample size

p=probability of success

q=probability of failure

C(n, x)=n!/(x!(n-x)!)

Therefore:

x=2

n=6

p=0.34

q=1-p=0.66

P(X=2)=C(6,2)*0.34²*0.66⁴

C(6,2)=6!/(2!*4!)

Do these calculations and you will have the probability of exactly two people winning free burgers. I hope this helps.