Question

A factory $A$ produces $10$% defective valves and another factory $B$ produces $20$% defective. A bag contains $4$ valves of factory $A$ and $5$ valves of factory $B$. If two valves are drawn at random from the bag. Find the probability that at least one valve is defective. Give your answer upto two places of decimals.

• A. $0.30$ approximately
• B. $0.28$ approximately
• C. $0.29$ approximately
• D. $0.27$ approximately

Answer 1

Answer: (C)

$0.29$ approximately

Probability of producing defective valves by factor $A=\frac{10}{100}=\frac{1}{10}$

Probability of producing defective valves by factory $B=\frac{20}{100}=\frac{1}{5}$

Bag $A$ contain $9$ valves, $4$ of factory $a$ and $5$ of factory $b$.

Two valves are $0$ be drawn at random.

$P$(at least one defective) $=$$1-P$(both are non defective)

$P$(both are non defective)$=$$P$(both valves of factory $B$)$\times P$(both are non defective) $+$$P$(both valves of factory $B$)$\times P$(both are non defective)$+$$P$(one valve of factory $A$ and other of factory $B$)$\times P$(Both are non defective)

$={\frac{{{}^{4}C}_2}{{}^9{C}_2}\left(\frac{9}{10}\right)}^2+\frac{{{}^{5}C}_2}{{}^9{C}_2}{\left(\frac{4}{5}\right)}^2+\frac{{}^4{C}_1{.}^5{C}_1}{{}^9{C}_2}.\frac{9}{10}.\frac{4}{5}$

$=\frac{1}{6}{\left(\frac{9}{10}\right)}^2+\frac{10}{36}.\frac{16}{25}+\frac{\mathrm{4.5.9.4}}{\mathrm{3.6.10.5}}$

$=\frac{27}{200}+\frac{8}{45}+\frac{2}{5}=\frac{1283}{1800}$

$\therefore P$(at least one defective) $=1-\frac{1283}{1800}=0.2872=0.29$(approx)