Question

A box contains $100$ bolts and $50$ nuts. It is given that $50\%$ bolts and $50\%$ nuts are rusted. Two objects are selected from the box at random. Find the probability that either both are bolts or both are rusted.

• A. $0.44$
• B. $0.58$
• C. $0.52$
• D. $0.48$

Answer 1

Answer: (B)

Total number of objects $=(100+50)=150$.
Let $S$ be the sample space. Then,
$n(S)=$ number of ways of selecting $2$ objects out of $150$
$={}^{150}{C}_2$.
Number of rusted objects
$=(50\%$ of $100)+(50\%$ of $50)=50+25=75$.
Let $E_1=$ event of selecting $2$ bolts out of $100$ bolts,
and $E_2=$ event of selecting $2$ rusted objects out of $75$ rusted objects.
$\therefore E_1\cap E_2=$ event of selecting $2$ rusted bolts out of $50$ rusted bolts.
$\therefore n(E_1)=$ number of ways of selecting $2$ bolts out of $100={}^{100}{C}_2$.
$\therefore n(E_2)=$ number of ways of selecting $2$ rusted objects out of $75={}^{75}{C}_2$.
$\therefore n(E_1\cap E_2)=$ number of ways of selecting $2$ rusted bolts out of $50={}^{50}{C}_2$.
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{{}^{100}{C}_2}{{}^{150}{C}_2}$,
$P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{{}^{75}{C}_2}{{}^{150}{C}_2}$
and $P\left(E_1\cap E_2\right)=\frac{n\left(E_1\cap E_2\right)}{n\left(S\right)}=\frac{{}^{50}{C}_2}{{}^{150}{C}_2}$.
$P$ (Either both are bolts or both are rusted)
$=P(E_1$ or $E_2)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1\cap E_2\right)$
$=\frac{{}^{100}{C}_2}{{}^{100}{C}_2}+\frac{{}^{75}{C}_2}{{}^{150}{C}_2}-\frac{{}^{50}{C}_2}{{}^{150}{C}_2}$
$=\frac{\left({}^{100}{C}_2{+}^{75}{C}_2-{}^{50}{C}_2\right)}{{}^{150}{C}_2}$
$=\frac{\left(4950+2775-1225\right)}{11175}$
$=\frac{6500}{11175}$
$=\frac{260}{447}$
$=0.58$.
Hence, the required probability is $0.58$.