Probability of choosing stale donuts out of 24

[TheSodesa]

Homework Statement

There are 6 stale donuts in a set of 24. What is the probability of:
a) there being no stale donuts in a sample of 10?
b) there being 3 stale donuts in a sample of 10?
c) What is the chance of a stale doughnut being found?

Homework Equations

N \, permutations = N!

The Attempt at a Solution

Let $X$ denote the number of stale donuts in a set of 10.

a) I used the idea of permutations like so:
P(X = 0) = \frac{\frac{18!}{8!}}{\frac{24!}{15!}} \approx 0.335
This was incorrect, according to the testing software

b) Here I followed the same idea:
P(X=3) = \frac{\frac{18!}{11!} \times \frac{6!}{3!}}{\frac{24!}{15!}} \approx 0.041

c) This is just the complement of part $a$:

P(X \, is \, at \, least \, 1) = 1 - 0.061 = 0.665

Parts $a$ and $b$ (and $c$ as a consequence of $a$ being wrong) are apparently wrong and I'm not sure what's wrong with my reasoning.

 

[TheSodesa]

I was completely wrong. I was supposed to use combinations instead:

a) P(X = 0) = \frac{18 \choose 10}{24 \choose 10} =39/1748 \approx 0.22

b) P(X = 3) = \frac{{18 \choose 7} \times {6 \choose 3}}{{24 \choose 10}} =1560/4807 \approx 0.325

c) P(X \text{ is at least } 1) = 1 - P(X = 0) = 1709/1748 \approx 0.987

 

[Buzz Bloom]

Hi TheSodesa:

You might find it to be of some interest that the general case of the problems you show involve the hypergeometric distribution:

https://en.wikipedia.org/wiki/Hypergeometric_distribution​

Regards,
Buzz

 

[Ray Vickson]

I was completely wrong. I was supposed to use combinations instead:

a) P(X = 0) = \frac{18 \choose 10}{24 \choose 10} =39/1748 \approx 0.22

b) P(X = 3) = \frac{{18 \choose 7} \times {6 \choose 3}}{{24 \choose 10}} =1560/4807 \approx 0.325

c) P(X \text{ is at least } 1) = 1 - P(X = 0) = 1709/1748 \approx 0.987

These are all correct, now. As Buzz Bloom has stated, you are using the so-called hypergeometric distribution.