Question

The probability that a screw manufactured by a company is defective is 0.1 The company sells screws in packets containing 5 screws and gives a guarantee of replacement if one or more screws in the packet are found to be defective. The probability that a packet would have to be replaced is _____

Answer 1

Concept:

The binomial probability formula:

Probability of r successes in n trials:

P(r) = nCrprqn-r

n = number of trials, r = number of success in n trials

p = probability of success in any one trial, q = probability of failure in any one trial = 1 – p

\({{\rm{P}}_{\left( {{\rm{r}},{\rm{n}}} \right)}} = {}_{}^{\rm{n}}{{\rm{C}}_{\rm{r}}}{{\rm{P}}_{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}{{\rm{P}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\)

Calculation:

The probability that a packet would have to be replaced =?

If P(r ≥ 1) packet have to le replaced

So, required probability is, 1 - P(r = 0)

\(\therefore P\left( {r =0} \right) = {^n{{C_o}}}{p^0}{q^{n - 0\;}}\)

Given, n = 5, p = 0.1, q = 1 - p = 0.9

\( \Rightarrow P\left( {r = 0} \right) = {^5{{C_0}}}{\left( {0.1} \right)^0}{\left( {0.9} \right)^5}\)

= 0.5905

So, required probability = 1 - P (X = 0) = 1 - 0.5905

P(r ≥ 1) = 0.4095