Question

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Answer 1

Let X = number of defective items

p = probability of defective item

∴ p = 5% = ${\displaystyle \frac{5}{100}=\frac{1}{20}}$

and q = 1 – p = ${\displaystyle 1-\frac{1}{20}=\frac{19}{20}}$

∴ ${\displaystyle X\sim B(10,\frac{1}{20})}$

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

P(X = x) = ${\displaystyle {}^n{C}_x p^x q^{n-x}}$

i.e. p(x) = ${\displaystyle {}^{10}{C}_x {\left(\frac{1}{20}\right)}^x{\left(\frac{19}{20}\right)}^{10-x}}$, x = 0, 1, 2, ..., 10

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

= P(X = 0) + P(X = 1)

= ${\displaystyle {}^{10}{C}_0{\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^{10-0}+{}^{10}{C}_1{\left(\frac{1}{20}\right)}^1{\left(\frac{19}{20}\right)}^{10-1}}$

= ${\displaystyle 1\cdot 1\cdot {\left(\frac{19}{20}\right)}^{10}+10\times \left(\frac{1}{20}\right)\times {\left(\frac{19}{20}\right)}^9}$

= ${\displaystyle {\left(\frac{19}{20}\right)}^9[\frac{19}{20}+\frac{10}{20}]}$

= ${\displaystyle {\left(\frac{19}{20}\right)}^9\left(\frac{29}{20}\right)}$

= ${\displaystyle 29\left(\frac{{19}^9}{{20}^{10}}\right)}$

Hence, the probability that a sample of 10 items will include not more than one defective item = ${\displaystyle 29\left(\frac{{19}^9}{{20}^{10}}\right)}$.