Exercise 1.58 (Widget inspector)

Question

A widget inspector inspects 12 widgets and finds that exactly 3 are defective. Unfortunately, the widgets then get all mixed up and the inspector has to find the 3 defective widgets again by testing widgets one by one.
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(a) Find the probability that the inspector will now have to test at least 9 widgets.
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(b) Find the probability that the inspector will now have to test at least 10 widgets.

fifthist x · Accepted Answer

Explanation: https://math.stackexchange.com/questions/1936525/inclusion-exclusion-problem
\begin{enumerate}[label=(\alph*)]
\item Let $A$ be the event that at least $9$ widgets need to be tested. 
$$P(A) = 1 - P(A^{c}) = 1 - \frac{\binom{8}{3}3!9!}{12!}$$

\item Similar to part $a$, $$P(A) = 1 - P(A^{c}) = 1 - \frac{\binom{9}{3}3!9!}{12!}$$
\end{enumerate}

Exercise 1.58 (Widget inspector)

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Exercise 1.58 (Widget inspector)

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