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Problem 97: A quality-control engineer inspects a random sample of 3 batteries from each lot of 24 car batteries ready to be shipped. If such a lot contains 6 batteries with slight defects, what probabilities that the inspector's sample will contain a. None of the batteries with defects? b. only one of the batteries with defects? c. At least two of the batteries with defects?

Problem 97: A quality-control engineer inspects a random sample of 3 batteries from each lot of 24 car batteries ready to be shipped. If such a lot contains 6 batteries with slight defects, what probabilities that the inspector's sample will contain a. None of the batteries with defects? b. only one of the batteries with defects? c. At least two of the batteries with defects?

Solution:


a. None of the batteries with defects?
Given that $x=0,n=3,a=6, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;3,6,24)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}24-6\\ 3-0\end{array}\right)}{\left(\begin{array}{c}24\\ 3\end{array}\right)} = 0.40316.$

b. only one of the batteries with defects?
Given that $x=1,n=3,a=6, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=1) = h(1;3,6,24)=\frac{\left(\begin{array}{c}6\\ 1\end{array}\right)\left(\begin{array}{c}24-6\\ 3-1\end{array}\right)}{\left(\begin{array}{c}24\\ 3\end{array}\right)} = 0.453557.$

c. At least two of the batteries with defects
Given that $x=1,n=3,a=6, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x\ge 2) = 1-P(x< 2) = 1- [P(x=0)+P(x=1)]$
$P(x=0) = h(0;3,6,24)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}24-6\\ 3-0\end{array}\right)}{\left(\begin{array}{c}24\\ 3\end{array}\right)} = 0.40316.$
$P(x=1) = h(1;3,6,24)=\frac{\left(\begin{array}{c}6\\ 1\end{array}\right)\left(\begin{array}{c}24-6\\ 3-1\end{array}\right)}{\left(\begin{array}{c}24\\ 3\end{array}\right)} = 0.453557.$
$P(x\ge 2) = 1- [P(x=0)+P(x=1)] = 1-(0.40316+0.453557) = 1- 0.8567 =0.1432$

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