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Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?


Solution:


a. none of the buildings that violate the building code?
Given that $x=0,n=4,a=6, N=18$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$

b. 1 of the new buildings that violate the building code?
Given that $x=1,n=4,a=6, N=18$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=1) = h(1;4,6,18)=\frac{\left(\begin{array}{c}6\\ 1\end{array}\right)\left(\begin{array}{c}18-6\\ 4-1\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.431373.$

c. 2 of the new buildings that violate the building code?
Given that $x=1,n=4,a=6, N=18$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=2) = h(2;4,6,18)=\frac{\left(\begin{array}{c}6\\ 2\end{array}\right)\left(\begin{array}{c}18-6\\ 4-2\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.323529.$

c. at least 3 of the new buildings that violate the building code
Given that $n=4,a=6, N=18$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x\ge 3) = 1-P(x< 3) = 1- [P(x=0)+P(x=1)+P(x=2)]$
$P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$
$P(x=1) = h(1;4,6,18)=\frac{\left(\begin{array}{c}6\\ 1\end{array}\right)\left(\begin{array}{c}18-6\\ 4-1\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.431373.$
$P(x=2) = h(2;4,6,18)=\frac{\left(\begin{array}{c}6\\ 2\end{array}\right)\left(\begin{array}{c}18-6\\ 4-2\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.323529.$
$P(x\ge 3) = 1- [P(x=0)+P(x=1)+P(x=2)] = 1-(0.16176+0.431373+0.323529) = 1- 0.916667=0.08333$

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