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Problem 83: Cumulative binomial probabilities

Problem 83: If the probability is 0.05 that a certain wide-flange column will fail under a given axial load, what are the probabilities that among 16 such columns


a . at most two will fail; b. at least four will fail?

Solution:


a. at most two will fail
Given that $n=5,$ and $p=0.6$
$\sum _{x=0}^{2}b(x;n,p)=\sum _{x=0}^{2}b(x;16,0.05)=b(0;16,0.05)+b(1;16,0.05)+b(2;16,0.05)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;16,0.05) = \left(\begin{array}{c}16\\ 0\end{array}\right){0.05}^{0}(1-0.05)^{16-0} = 0.440127$
$b(1;16,0.05) = \left(\begin{array}{c}16\\ 1\end{array}\right){0.05}^{1}(1-0.05)^{16-1} = 0.370633$
$b(2;16,0.05) = \left(\begin{array}{c}16\\ 2\end{array}\right){0.05}^{2}(1-0.05)^{16-1}=0.146302$
$b(0;16,0.05)+b(1;16,0.05)+b(2;16,0.05) = 0.440127+0.370633+0.146302 = 0.957062.$
probability at most two will fail is 0.957062. b. at least four will fail
Given that $n=5,$ and $p=0.6$
$\sum _{x=4}^{16}b(x;n,p)=\sum _{x=4}^{16}b(x;16,0.05)=1-\sum _{x=0}^{3}b(x;16,0.05) = 1- [b(0;16,0.05)+b(1;16,0.05)+b(2;16,0.05)+b(3;16,0.05)]$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;16,0.05) = \left(\begin{array}{c}16\\ 0\end{array}\right){0.05}^{0}(1-0.05)^{16-0} = 0.440127$
$b(1;16,0.05) = \left(\begin{array}{c}16\\ 1\end{array}\right){0.05}^{1}(1-0.05)^{16-1} = 0.370633$
$b(2;16,0.05) = \left(\begin{array}{c}16\\ 2\end{array}\right){0.05}^{2}(1-0.05)^{16-2}=0.146302$
$b(3;16,0.05) = \left(\begin{array}{c}16\\ 3\end{array}\right){0.05}^{3}(1-0.05)^{16-3}=0.035934$
$\sum _{x=4}^{16} = 1- [b(0;16,0.05)+b(1;16,0.05)+b(2;16,0.05)+b(3;16,0.05)] = 1- (0.440127+0.370633+0.146302+0.035934) = 1-0.993 = 0.0070.$
probability at least four will fail is 0.0070.

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