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

The correct option is C

$\frac{11}{16}$

If the numbers of nails and nuts are 6 and 10, respectively, then the numbers of rusted nails and rusted nuts are 3 and 5, respectively.

Total number of items $=6+10=16$

Total number of rusted items $=3+5=8$

Total number of ways of drawing one item ${=}^{16}{C}_1$

Let R and N be the events where both the items drawn are rusted items and nails, respectively.

R and N are not mutually exclusive events, becuase there are 3 rusted nails.

P(either a rusted item or a nail)

$=P(R\cup N)$

$=P\left(R\right)+P\left(N\right)-(R\cap N)$

$=\frac{{}^6{C}_1}{{}^{16}{C}_1}+\frac{{}^6{C}_1}{{}^{16}{C}_1}-\frac{{}^3{C}_1}{{}^{16}{C}_1}$

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