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

Options

•  3/16

•  5/16

•  11/16

•  14/16

   

Answer 1

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 = ¹⁶C₁
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, because there are 3 rusted nails.
P(either a rusted item or a nail ) = P (R ∪ N)
                                                = P(R) + P (N) - P (R ∩ N)
                                                 =\[\frac{^{6}{}{C}_1}{^{16}{}{C}_1} + \frac{^{8}{}{C}_1}{^{16}{}{C}_1} - \frac{^{3}{}{C}_1}{^{16}{}{C}_1}\]

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