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Problem 104: Suppose that, next month, the quality control division will inspect 30 units. Among these, 20 will undergo a speed test and 10 will be tested for current flow. If an engineer is randomly assigned 4 units, what are the probabilities that
a. none of them will need a speed test?
b. only 2 will need a speed test?
c. at least 3 will need a speed test?

Problem 104: Suppose that, next month, the quality control division will inspect 30 units. Among these, 20 will undergo a speed test and 10 will be tested for current flow. If an engineer is randomly assigned 4 units, what are the probabilities that
a. none of them will need a speed test?
b. only 2 will need a speed test?
c. at least 3 will need a speed test?


Solution:


a. none of them will need a speed test
Given that $x=0,n=4,a=20, N=30$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;4,20,30)=\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}30-20\\ 4-0\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.007663.$

b. only 2 will need a speed test
Given that $x=0,n=4,a=6, N=13$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=2) = h(2;4,20,30)=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}30-20\\ 4-2\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.311987.$

c. at least 3 will need a speed test
Given that $n=4,a=6, N=13$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x\ge 3) = P(x=3) + P(x=4)$
$P(x=3) = h(3;4,20,30)=\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}30-20\\ 4-3\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.41592.$

$P(x=4) = h(4;4,20,30)=\frac{\left(\begin{array}{c}20\\ 4\end{array}\right)\left(\begin{array}{c}30-20\\ 4-4\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.1767.$

$P(x\ge 3) = P(x=3) + P(x=4) = 0.41592+0.1767 = 0.5927$

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