Question

A box contains $6$ nails and $10$ nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or is a nail?

Answer 1

Let $A$ be the event that the item chosen is rusted and $B$ be the event that the them chosen is a nail.

$\therefore P\left(A\right)=\frac{8}{16}$

$P\left(B\right)=\frac{6}{16}$

and $P(A\cap B)=\frac{3}{16}$

Required probability$=P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}=\frac{8+6-3}{16}=\frac{11}{16}$