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Problem 88: During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chip receive a thick enough coating, find the probability that

Problem 88: During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chip receive a thick enough coating, find the probability that, among 15 chip, (a) at least 12 will have thick enough coatings; (b) at most 6 will have thick enough coatings; (c) exactly 10 will have thick enough coating.

Solution:


a. at least 12 will have thick enough coatings


Given that $n=15,$ and $p=0.7$
$P(x\ge 12) = \sum _{x=12}^{15}b(x;n,p)=\sum _{x=12}^{15}b(x;15,0.7)$
$=b(12;15,0.7)+b(13;15,0.7)+b(14;15,0.7)+b(15;15,0.7)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(12;15,0.7) = \left(\begin{array}{c}15\\ 12\end{array}\right){0.7}^{12}(1-0.7)^{15-12} = 0.17004 $
$b(13;15,0.7) = \left(\begin{array}{c}15\\ 13\end{array}\right){0.7}^{13}(1-0.7)^{15-13} = 0.09156$
$b(14;15,0.7) = \left(\begin{array}{c}15\\ 14\end{array}\right){0.7}^{14}(1-0.7)^{15-14} = 0.03052$
$b(15;15,0.7) = \left(\begin{array}{c}15\\ 15\end{array}\right){0.7}^{15}(1-0.7)^{15-15} = 0.004748$
$P(x\ge 12) = b(12;15,0.7)+b(13;15,0.7)+b(14;15,0.7)+b(15;15,0.7) $
$= 0.17004+0.09156+0.03052+0.004748 = 0.2968.$
b. at most 6 will have thick enough coatings

Given that $n=15,$ and $p=0.7$
$P(x\le 6) = \sum _{x=0}^{6}b(x;n,p)=\sum _{x=0}^{6}b(x;15,0.7)$
$=b(0;15,0.7)+b(1;15,0.7)+b(2;15,0.7)+b(3;15,0.7)+b(4;15,0.7)+b(5;15,0.7)+b(6;15,0.7)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;15,0.7) = \left(\begin{array}{c}15\\ 0\end{array}\right){0.7}^{0}(1-0.7)^{15-0} = 1.43x10^{-08} $
$b(1;15,0.7) = \left(\begin{array}{c}15\\ 1\end{array}\right){0.7}^{1}(1-0.7)^{15-1} = 5.02x10^{-07}$
$b(2;15,0.7) = \left(\begin{array}{c}15\\ 2\end{array}\right){0.7}^{2}(1-0.7)^{15-2} = 8.2x10^{-06}$
$b(3;15,0.7) = \left(\begin{array}{c}15\\ 3\end{array}\right){0.7}^{3}(1-0.7)^{15-3} = 8.29x10^{-05}$
$b(4;15,0.7) = \left(\begin{array}{c}15\\ 4\end{array}\right){0.7}^{4}(4-0.7)^{15-4} = 0.000581$
$b(5;15,0.7) = \left(\begin{array}{c}15\\ 5\end{array}\right){0.7}^{5}(4-0.7)^{15-5} = 0.00298$
$b(6;15,0.7) = \left(\begin{array}{c}15\\ 6\end{array}\right){0.7}^{6}(4-0.7)^{15-6} = 0.01159$
$P(x\le 6) = b(0;15,0.7)+b(1;15,0.7)+b(2;15,0.7)+b(3;15,0.7)+b(4;15,0.7)+b(5;15,0.7)+b(6;15,0.7) $
$= 0.0152.$
c. exactly 10 will have thick enough coating
Given that $n=15,$ and $p=0.7$
$P(x= 10) = b(10;15,0.7)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x= 10) = b(10;15,0.7) = \left(\begin{array}{c}15\\ 10\end{array}\right){0.7}^{10}(4-0.7)^{15-10} = 0.20613.$

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