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, then find the probability that it is rusted or is a nail.

• A. $\frac{11}{16}$
• B. $\frac{3}{16}$
• C. $\frac{5}{16}$
• D. $\frac{13}{16}$

Answer 1

Answer: (A)

$\frac{11}{16}$

The total number of nails and nuts is $6+10=16$.
$P\left(R\right)=$ $\frac{1}{2}$ ( R stands for rusted)

$P\left(N\right)=$ $\frac{6}{16}$ ( $N$ stands for nails)

$P(R\cap N)=\frac{3}{16}$

[$\because$ $3$ nails are rusted out of $6$ nails]
$P(R\cup N)=P\left(R\right)+P\left(N\right)-P(P\cap N)$$=\frac{1}{2}+\frac{6}{16}-\frac{3}{16}$$=\frac{8+6-3}{16}=\frac{11}{16}$