Question

A manufacturing company produces batteries, the quality control manager found the following facts. A battery is either good or bad. Based on the previous record, the probability of selecting a bad battery is 0.04. The probability of producing each battery is independent of others.
a) If we randomly select 10 batteries, what is the probability that exactly 7 of them are good.
b) If we randomly select 15 batteries, what is the probability that at least 13 of them are bad.
c) If we randomly select 20 batteries, what is the probability that at most 18 of them are good.

Answer 1

The probability of selecting a bad battery is $q=0.04$ and the probability of selecting a good battery is $p=1-q=0.96$ .

We can consider this process as binomial experiment:

suppose we select $n$ batteries and let $X$ be the number of good batteries.

Then $X\sim B(n,p)$ . We will use formula $P(X=k)= \binom{n}{k} p^k q^{n-k}$ .

a) $n=10$

$P(X=7)= \binom{10}{7} 0.96^7 \cdot 0.04^{3}=120\cdot 0.96^7 \cdot 0.04^{3}=0.0057$

b) $n=15$

The probability that at least 13 batteries are bad is equal to the probability that at most 2 batteries are good.

$P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)= \\ =\binom{15}{0} 0.96^0\cdot 0.04^{15}+ \binom{15}{1} 0.96^1\cdot 0.04^{14}+ \binom{15}{2} 0.96^2\cdot 0.04^{13}=\\ = 1\cdot 0.04^{15}+ 15\cdot 0.96\cdot 0.04^{14}+ 105\cdot 0.96^2\cdot 0.04^{13}\approx 0$

c) $n=20$

$P(X\leq 18)=1-P(20\geq X\geq 19)=1-P(X=19)-P(X=20)= \\ =1-\binom{20}{20} 0.96^{20}\cdot 0.04^{0}- \binom{20}{19} 0.96^{19}\cdot 0.04^{1}=\\ = 1-1\cdot 0.96^{20}-20\cdot 0.96^{19}\cdot 0.04\approx 0.18966$

Answer: a) 0.0057; b) 0; c) 0.18966.