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, the probability that it is rusted or is a nail is

(a) $\frac{3}{16}$

(b) $\frac{5}{16}$

(c) $\frac{11}{16}$

(d) $\frac{14}{16}$

Answer 1

(c) $\frac{11}{16}$


$$\begin{gathered} \text{Rusted items =3+5=8} \\ \mathrm{Rusted}\mathrm{nails}=3 \\ \mathrm{Total}\mathrm{nails}=6 \\ P\left(\mathrm{getting}\mathrm{a}\mathrm{rusted}\mathrm{item}\mathrm{or}\mathrm{a}\mathrm{nail}\right)=P\left(\mathrm{getting}\mathrm{a}\mathrm{rusted}\mathrm{item}\right)+P\left(\mathrm{getting}\mathrm{a}\mathrm{nail}\right)-P\left(\mathrm{getting}\mathrm{a}\mathrm{rusted}\mathrm{item}\mathrm{and}\mathrm{a}\mathrm{nail}\right) \\ =\frac{8}{16}+\frac{6}{16}-\frac{3}{16} \\ =\frac{8+6-3}{16} \\ =\frac{11}{16} \end{gathered}$$