Answer to Question #152028 in Statistics and Probability for JG

Question

A manufacturing company produces computer chips, the quality control manager found the following facts.
1. A chip is either good or bad.
2. Based on the previous record, the probability of selecting a good chip is 0.92.
3. The probability of producing each chip is independent of others.

a) If we randomly select 10 chips, what is the probability that exactly 8 of them are good.
b) If we randomly select 15 chips, what is the probability that no more than 2 of them are bad.
c) If we randomly select 20 chips, what is the probability that at least 18 of them are good.

Answer 1

Here x follows the binomial distribution

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}$

a) n = 10

p = 0.92

$P(X=8)=C(10,8)\cdot 0.92^8\cdot (1-0.92)^{10-8}=\frac{10!}{8!2!}\cdot 0.92^8\cdot0.08^2=0.1478$

b) n = 15

p = 1 – 0.92 = 0.08

$P(X\le2)=P(X=0)+P(X=1)+P(X=2)$

$P(X=0)=C(15,0)\cdot0.08^0\cdot0.92^{15}=0.2863$

$P(X=1)=C(15,1)\cdot0.08^1\cdot0.92^{14}=0.3734$

$P(X=2)=C(15,2)\cdot0.08^2\cdot0.92^{13}=0.2273$

$P(X\le2)=0.2863+0.3734+0.2273=0.887$

c) n = 20

p = 0.92

$P(X\ge18)=P(X=18)+P(X=19)+P(X=20)$

$P(X=18)=C(20,18)\cdot0.92^{18}\cdot0.08^{2}=0.2711$

$P(X=19)=C(20,19)\cdot0.92^{19}\cdot0.08^{1}=0.3281$

$P(X=20)=C(20,20)\cdot0.92^{20}\cdot0.08^{0}=0.1887$

$P(X\ge18)=0.2711+0.3281+0.1887=0.7879$

Answer:

a) 0.1478

b) 0.887

c) 0.7879