Osvaldo Apodaca

2022-01-18

Products from a certain machine are too large 15% of the time. What is the probability that in a run of 20 parts, 5 are too large?

Edward Patten

Expert

${C}_{5}^{20}{\left(.15\right)}^{5}{\left(.85\right)}^{15}\approx .103$
Explanation:
The probability that 5 parts are too large is $.15×.15×.15×.15×.15={\left(.15\right)}^{5}$
and since 5 parts are too large, the other 15 are of acceptable size with probability
${\left(1-.15\right)}^{15}.$ Since any 5 parts can be too large in no particular order, there are ${C}_{5}^{20}$
ways to do this, so the probability that exactly 5 parts are too large is
${C}_{5}^{20}{\left(.15\right)}^{5}{\left(.85\right)}^{15}\approx .103.$

redhotdevil13l3

Expert

Given, product from a certain machine too large
is 15% of time i.e p=0.15
$\therefore$ q=1-p=1-0.15=0.85
sample size (n)=20
Binomial distribution: $P\left(X=\gamma \right)={n}_{{C}_{\gamma }}{p}^{\gamma }{q}^{n-\gamma }$
$\therefore p\left(X=5\right)={20}_{{C}_{5}}{\left(0.15\right)}^{5}{\left(0.85\right)}^{15}=0.1028$