A box contains 6 nails and 10 nuts. Half of t | vedclass

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(C) Total number of items $= 6 + 10 = 16$.Let $N$ be the event that the item is a nail and $R$ be the event that the item is rusted.Number of nails $n

Vedclass · Accepted Answer

$\frac{11}{16}$

Vedclass · Answer

(C) Total number of items $= 6 + 10 = 16$.Let $N$ be the event that the item is a nail and $R$ be the event that the item is rusted.Number of nails $n(N) = 6$.Number of rusted items $n(R) = \frac{6}{2} + \frac{10}{2} = 3 + 5 = 8$.Number of rusted nails $n(N \cap R) = \frac{6}{2} = 3$.We need to find the probability of $N \cup R$.Using the formula $P(N \cup R) = P(N) + P(R) - P(N \cap R)$:$P(N \cup R) = \frac{6}{16} + \frac{8}{16} - \frac{3}{16} = \frac{11}{16}$.

$A$ 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,what is the probability that it is rusted or is a nail?

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