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) 3/16
(b) 5/16
(c) 11/16
(d) 14/16

Answer 1

A box contain 6 nails and 10 nuts.

Let A be the event of drawing a nail
i.e. P(A)$=\frac{6}{16}$
Number of rusted nails = 3, number of rusted nuts = 5
Let B the event of drawing rusted item$=\frac{8}{16}$
Then A∩B defines an event of drawing rusted nail = $\frac{3}{16}$
Then P(A∪B) = P(A) + P(B) − P(A∩B)
$$\begin{gathered} =\frac{6}{16}+\frac{8}{16}-\frac{3}{16} \\ \mathrm{P}(A\cup B)=\frac{11}{16} \end{gathered}$$
i.e. probability of drawing rusted item or nail = $\frac{11}{16}$
Hence, the correct answer is option C.