Question

A factory received a shipment of 40 capacitors, and the vendor who sold the items knows there are 7 capacitors in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the capacitors in the sample are defective, he will refuse the shipment.

For each of the following, give your responses as reduced fractions.

If a sample of 7 capacitors is selected, find the probability that all in the sample are defective.

  

If a sample of 7 capacitors is selected, find the probability that none in the sample are defective.

Answer 1

Solution-:

Let, X= Number of  defective capacitors in the shipment among 40 capacitors

Given:

$\therefore p=\frac{X}{n}=\frac{7}{40}=0.1750$

This situation is suitable for binomial distribution with parameter $n=40$   and  $p=0.1750$

$\therefore X\rightarrow B(n=40,p=0.1750)$

The p.m.f. of X is given by,

$P(X=x)=\binom{40}{x}*(0.1750)^x*(1-0.1750)^{40-x}$

$=0$   

(1) If a sample of 7 capacitors is selected, find the probability that all in the sample are defective.

P[all in the sample are defective] $=P[X=7]$

$=\binom{40}{7}*(0.1750)^7*(1-0.1750)^{40-7}$

$=0.1640$

The required probability is 0.1640

(2) If a sample of 7 capacitors is selected, find the probability that none in the sample are defective.

P[none in the sample are defective.] $=P[X=0]$

$=\binom{40}{0}*(0.1750)^0*(1-0.1750)^{40-0}$

$=0.0005$

  The required probability is 0.0005