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TheSodesa
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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
[itex]N \, permutations = N![/itex]
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:
[tex]P(X = 0) = \frac{\frac{18!}{8!}}{\frac{24!}{15!}} \approx 0.335[/tex]
This was incorrect, according to the testing software
b) Here I followed the same idea:
[tex]P(X=3) = \frac{\frac{18!}{11!} \times \frac{6!}{3!}}{\frac{24!}{15!}} \approx 0.041[/tex]
c) This is just the complement of part ##a##:
[tex]P(X \, is \, at \, least \, 1) = 1 - 0.061 = 0.665[/tex]
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.
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