Question & Answer
QUESTION

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?
(a) \[\dfrac{3}{8}\]
(b) $\dfrac{1}{4}$
(c) $\dfrac{3}{4}$
(d) $\dfrac{5}{8}$

ANSWER Verified Verified
Hint: Let A be the event that the item chosen is rusted. Let B be the event that the item chosen is a bolt. Find P(A) and P(B). Next, find $P(A\cap B)$ using the fact that A and B are independent events and the probability of intersection of two independent events is the product of the probability of individual events. Next, find \[P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)\] which is our final answer.



Complete step-by-step answer:
In this question, we are given that 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, we need to find the probability that it is rusted or a bolt.
Let A be the event that the item chosen is rusted. Clearly, there are 200 items in all, out of which 100 are rusted.
Therefore, $P\left( A \right)=\dfrac{100}{200}=\dfrac{1}{2}$
Let B be the event that the item chosen is a bolt. Clearly, there are 200 items in all, out of which 50 are bolts.
Therefore, $P\left( B \right)=\dfrac{50}{200}=\dfrac{1}{4}$
We observe that these two events are independent of each other.
We know that the probability of intersection of two independent events is the product of the probability of individual events.
Using this property, we get the following:
$P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)=\dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{1}{8}$
Now, in the question, we are to find the probability that the chosen item is rusted or a bolt.
So, we need to find \[P\left( A\cup B \right)\]
We know the formula for union that:
\[P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)\]
Using this formula, we will get the following:
\[P\left( A\cup B \right)=\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{5}{8}\]
So, the probability that the chosen item is rusted, or a bolt is $\dfrac{5}{8}$.
Hence, option (d) is correct.

Note: In this question, it is very important to know the formula for probability:
The probability of an event = $\dfrac{\text{No}\text{. of favourable outcomes}}{\text{Total no}\text{. of outcomes}}$. Without this formula, you will not be able to solve the question. Also, you need to know that the probability of intersection of two independent events is the product of the probability of individual events.