Question

A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random from the box, what is the probability that either both are rusted or both are bolts?

Answer 1

The number of bolts and nuts are 30 and 40 respectively. The number of rusted bolts and rusted nuts are $15(=\frac{30}{2})$ and $20(=\frac{40}{2})$ respectively.

Total number of items $=30+40=70$

Total number of rusted items $=15+20=35$

Total number of ways of drawing 2 items = ${}^{70}{C}_2$.

Let R and B be the events that the both item drawn are rusted items and both are bolts respectively.

R and B are not mutually exclusive events because there are 15 rusted bolts.

$\therefore$ P (items are both rusted or both bolts) $=P(R\cup B)=P\left(R\right)+P\left(B\right)-P(R\cap B)$

$=\frac{{}^{35}{C}_2}{{}^{70}{C}_2}+\frac{{}^{30}{C}_2}{{}^{70}{C}_2}-\frac{{}^{15}{C}_2}{{}^{70}{C}_2}$

$=\frac{1\times 2}{70\times 69}(\frac{35\times 34}{1\times 2}+\frac{30\times 29}{1\times 2}-\frac{15\times 14}{1\times 2})$

$=\frac{1850}{70\times 69}=\frac{185}{483}$