Probability of choosing stale donuts out of 24

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1. Sep 2, 2016

TheSodesa

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$N \, permutations = N!$

3. 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 whats wrong with my reasoning.

Last edited: Sep 2, 2016
2. Sep 2, 2016

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

3. Sep 2, 2016

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:

Regards,
Buzz

4. Sep 2, 2016

Ray Vickson

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