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 $0.58$.