A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is
A$\frac{3}{16}$
B$\frac{5}{16}$
C$\frac{11}{16}$
D$\frac{14}{16}$
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C$\frac{11}{16}$
Rusted items = 3 + 5 = 8
Rusted nails = 3
Total nails = 6
P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)
$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$
$=\frac{8+6-3}{16}$
$=\frac{11}{16}$
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