Question

A box has 50 pens of which 20 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement, at most one is defective?

• A. $3.6\times {\left(\frac{3}{5}\right)}^4$
• B. ${\left(\frac{3}{5}\right)}^5$
• C. $2.6\times {\left(\frac{3}{5}\right)}^4$
• D. ${\left(\frac{2}{5}\right)}^5$

Answer 1

The correct option is C $2.6\times {\left(\frac{3}{5}\right)}^4$

$\mathrm{Clearly},\mathrm{number}\mathrm{of}\mathrm{defective}\mathrm{pens}\mathrm{is}a\mathrm{random}\mathrm{variable}.\mathrm{Let}\mathrm{it}\mathrm{be}\mathrm{denoted}\mathrm{by}X.\mathrm{Let}p\mathrm{denote}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}a\mathrm{defective}\mathrm{pen}.\mathrm{Then},p=\frac{20}{50}=\frac{2}{5}\mathrm{and}q=1-\frac{2}{5}=\frac{3}{5}\mathrm{For}\mathrm{selecting}\mathrm{at}\mathrm{most}\mathrm{one}\mathrm{defective}\mathrm{pen},\mathrm{we}\mathrm{have}\mathrm{two}\mathrm{conditions},\mathrm{either}\mathrm{no}\mathrm{pen}\mathrm{is}\mathrm{defective}\mathrm{or}\mathrm{only}\mathrm{one}\mathrm{pen}\mathrm{is}\mathrm{defective}.\mathrm{The}\mathrm{problem}\mathrm{is}a\mathrm{binomial}\mathrm{experiment},\mathrm{with}n=5,p=\frac{2}{5}\mathrm{and}q=\frac{3}{5}.\mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{at}\mathrm{most}\mathrm{one}\mathrm{defective}\mathrm{pen}\mathrm{is}P\left(X=0\right)+P\left(X=1\right).\mathrm{or},5C0{\left(\frac{2}{5}\right)}^0\times {\left(\frac{3}{5}\right)}^5+5C1{\left(\frac{2}{5}\right)}^1\times {\left(\frac{3}{5}\right)}^4\mathrm{or},1\times 1\times {\left(\frac{3}{5}\right)}^5+5\times \frac{2}{5}\times {\left(\frac{3}{5}\right)}^4\mathrm{or},{\left(\frac{3}{5}\right)}^4\left(\frac{3}{5}+2\right)\mathrm{or},{\left(\frac{3}{5}\right)}^4\left(\frac{13}{5}\right)\mathrm{or},2.6\times {\left(\frac{3}{5}\right)}^4$