A drawer contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or a bolt ?

\frac{3}{4}

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\frac{5}{8}

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\frac{4}{7}

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\frac{5}{7}

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Solution

The correct option is C $\frac{5}{8}$
Let A be the event that the item chosen is rusted and B be the event that the item chosen is a bolt. Clearly, there are 200 items in all, out of which 100 are rusted.
Therefore,
$P (A) = \frac{100}{200}, P (B) = \frac{50}{200}, P (A \cap B) = \frac{25}{200}$

Required Probability $= P (A \cup B)$
$= P (A) + P (B) - P (A \cap B) = \frac{100}{200} + \frac{50}{200} - \frac{25}{200} = \frac{125}{200} = \frac{5}{8}$

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