# Thread: Basic Binomial Distribution... Complex Permutation

1. ## Basic Binomial Distribution... Complex Permutation

Hey guys, here's my problem:

"Ten percent of cadets at the police academy chew gum regularly. What is the probability (to the nearest percent) that, if you pick 4 cadets at random, two of them chew gum regularly?"

Well, that was easy. You use the binomial distribution. Part 2 of the question gets tricky though.

"If you 800 of them, what is the probability that 83 chew gum regularly? (answer is more than 37%)"

Here is my equation:

800 C 83 * 0.10^83(1 - 0.10)^717

Some pretty big, and small numbers that make my calculator hate me.

The permutation isn't hard to understand, just big. I understand that you will only have 800! to 718!, but there's got to be a way of computing that other than long hand writing it down.

Also, what about the tiny numbers?

thanks!

2. Originally Posted by Jasonvdm
Hey guys, here's my problem:

"Ten percent of cadets at the police academy chew gum regularly. What is the probability (to the nearest percent) that, if you pick 4 cadets at random, two of them chew gum regularly?"

Well, that was easy. You use the binomial distribution. Part 2 of the question gets tricky though.

"If you 800 of them, what is the probability that 83 chew gum regularly? (answer is more than 37%)"

Here is my equation:

800 C 83 * 0.10^83(1 - 0.10)^717

Some pretty big, and small numbers that make my calculator hate me.

The permutation isn't hard to understand, just big. I understand that you will only have 800! to 718!, but there's got to be a way of computing that other than long hand writing it down.

Also, what about the tiny numbers?

thanks!
The formula you wrote down is for exactly 83 chewing gum, whereas answer given in parentheses is for at least 83 chewing gum (but actual percentage is closer to 38%).

I think we generally use approximation, either normal or Poisson (be sure to check conditions for when these provide good approximations)

Binomial distribution - Wikipedia, the free encyclopedia

although it is possible to compute directly if you have the right tools, such as Pari/GP. (In fact, using Pari/GP you can get the answer to this question as rational p/q in lowest terms. )

3. Sorry, the question does want "at least" i guess i didn't type that out correctly. The answer book (i'm homeschooled) according to my mom, talks about z values. Could I really work this out by using standard deviation?

4. When you use the normal approximation, I get the probability as being 38.4%.

When you do the approximation to normal, you get:

X ~ N(np, npq)

where n is the sample size; 800
p the probability of chewing gum; 10%
and q the probability of not chewing gum; 90%

So, X~N(80, 72)

Can you take it from there?

5. Hmmm, i don't think the textbook is using the Poisson formula for this question. With the Poisson, you're not talking about a specific population. What are you using to calculate your answer?

thanks!

6. Sorry, I didn't notice your post before I posted.

I'm using the normal approximation.

$P(X > 83) = P(X > 82.5)$ -(continuity correction)

$z = \frac{x-\mu}{\sigma} = \frac{82.5-80}{\sqrt{72}} = 0.295$

Then, I look in my z table to find the associated probability.

7. that makes sense! Thanks a bunch!!