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Problem 102: Amaker of specialized instruments receives shipments of 24 circuit boards. Suppose one shipment contains 4 that are defective. An engineer selects a random sample of size 4. What are the probabilities that the sample will contain
a. 0 defective circuit boards?
b. 1 defective circuit board ?
c. 2 or more defective circuit boards?

Problem 102: Amaker of specialized instruments receives shipments of 24 circuit boards. Suppose one shipment contains 4 that are defective. An engineer selects a random sample of size 4. What are the probabilities that the sample will contain
a. 0 defective circuit boards?
b. 1 defective circuit board ?
c. 2 or more defective circuit boards?


Solution:


a. 0 defective circuit boards
Given that $x=0,n=4,a=4, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;4,4,24)=\frac{\left(\begin{array}{c}4\\ 0\end{array}\right)\left(\begin{array}{c}24-4\\ 4-0\end{array}\right)}{\left(\begin{array}{c}24\\ 4\end{array}\right)} = 0.456.$

b. 1 defective circuit board
Given that $x=1,n=4,a=4, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=1) = h(1;4,4,24)=\frac{\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}24-4\\ 4-1\end{array}\right)}{\left(\begin{array}{c}24\\ 4\end{array}\right)} = 0.4291.$

c. 2 or more defective circuit boards
Given that $n=4,a=4, N=24$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x\ge 2) = 1-P(x<2) = 1-[P(x=0)+P(x=1)]$
$P(x=0) = h(0;4,4,24)=\frac{\left(\begin{array}{c}4\\ 0\end{array}\right)\left(\begin{array}{c}24-4\\ 4-0\end{array}\right)}{\left(\begin{array}{c}24\\ 4\end{array}\right)} = 0.456$
$P(x=1) = h(1;4,4,24)=\frac{\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}24-4\\ 4-1\end{array}\right)}{\left(\begin{array}{c}24\\ 4\end{array}\right)} = 0.4291$
$P(x\ge 2) = 1-[P(x=0)+P(x=1)] = 1-(0.456+0.4291) = 1- 0.88509 = 0.1149$

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