Problem 84: If the probability is 0.3 that an LCD panel will not pass inspection, what is the probability that 6 of 18 panels, randomly selected from production, will not pass inspection?

Problem 84: If the probability is 0.3 that an LCD panel will not pass inspection, what is the probability that 6 of 18 panels, randomly selected from production, will not pass inspection?

Solution:

Given that $x=6,n=18,$ and $p=0.3$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(6;18,0.3) = \left(\begin{array}{c}18\\ 6\end{array}\right){0.3}^{6}(1-0.3)^{18-6}$
$^{18}C_{6}{0.3}^{6}(0.7)^{12} = 0.1873$