Probability binomial distribution

Question

The probability of buying a movie ticket with popcorn is 0.761, and the probability of buying a movie ticket without popcorn is 0.239. If you buy 24 movie tickets we want to know the probability that more than 16 of the tickets have popcorn. What does p=?

Answer 1

Hi Tiffany,

This is a LONG calculation, just so you're prepared. We use the binomial probability formula here. This is appropriate to use whenever you have a binary outcome and trials are independent. Here, you have a binary outcome--either popcorn or not--and trials are theoretically independent--one person buying popcorn won't influence anyone else's buying popcorn, so we can proceed with the formula:

P=C(n,x) p^xq^n-x

Let's break this down:

P=probability we are looking for

C=notation that we have a combination, order of purchase does not matter

n=total customers

x=customers who bought popcorn

p=probability that customer bought popcorn

q=1-p, probability that customer didn't buy popcorn

Now, C(n, x) is the longest part of the calculation. It's read as "n choose x." In this case, it's 24, choose 16. The formula to compute this count is:

C(n, x)=n!/(x!*(n-x)!) where:

!=factorial=n*(n-1)*(n-2)... i.e. 3*2*1, etc.

Here,

n=24

x=16

C(24,16)=24!/(16!*8!)=735471

You can input this into a graphing calculator like the TI-83 series.

Now, we can plug this back into our original formula:

P=C(n,x) p^xq^n-x

C(n,x)=735471

n=24

x=16

p=0.761

q=1-0.761=0.239

Substituting:

P₁₆=735471*0.761¹⁶*0.239⁸

P₁₆=0.099

But we're far from done. That was just the probability that exactly 16 of 24 people buy popcorn, not the probability that 16 or greater buy popcorn. To get this, we have to repeat the formula:

P=C(n,x) p^xq^n-x

for x=17, 18, 19, 20, 21, 22, 23, 24

All of these calculations are done the same way--just sub each value in for x. Once you complete that, add up all of the probabilities beginning with the 0.099 we computed above. I got approximately:

P=0.099 + 0.148 + 0.184 + 0.185 + 0.147 + 0.089 + 0.039 + 0.011 + 0.001=

P=0.903

While binomial probability problems are certainly fair game in statistics courses, I hope they do not give you more with this many calculations in the future! I hope this helps.