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

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(C) Total number of items in the box $= 6 + 10 = 16$.Number of rusted nails $= \frac{1}{2} 	imes 6 = 3$.Number of rusted nuts $= \frac{1}{2} 	imes 1

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$\frac{11}{16}$

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(C) Total number of items in the box $= 6 + 10 = 16$.Number of rusted nails $= \frac{1}{2} 	imes 6 = 3$.Number of rusted nuts $= \frac{1}{2} 	imes 10 = 5$.Total number of rusted items $= 3 + 5 = 8$.Number of non-rusted nails $= 6 - 3 = 3$.We need to find the probability that the chosen item is rusted or is a nail.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:Let $A$ be the event that the item is rusted,and $B$ be the event that the item is a nail.$P(A) = \frac{8}{16}$,$P(B) = \frac{6}{16}$,and $P(A \cap B) = \frac{3}{16}$ (rusted nails).Probability $= \frac{8}{16} + \frac{6}{16} - \frac{3}{16} = \frac{11}{16}$.

$A$ 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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