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A garage mechanic keeps a box of good springs to use as replacements on customers' cars. The box contains $5$ springs. A colleague, thinking that the springs are for scrap, tosses three faulty springs into the box. The mechanic picks the two springs out of the box while servicing a car. Find the probability that the second spring drawn is faulty.A. $\dfrac{1}{8}$B. $\dfrac{2}{8}$C. $\dfrac{3}{8}$D. $\dfrac{4}{8}$

Last updated date: 09th Aug 2024
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Hint:Note down the number of good springs, faulty springs and the total number of springs in the box. Find the probability for the first spring to be chosen faulty and also the probability of the second spring to be chosen faulty. Now apply the Baye’s theorem as there is an event happening prior to the main event.

In order to solve this problem, firstly let us know about the most important theorem in probability which we are going to use to solve this.
The name of the theorem is Bay’s theorem and it is defined as follows:
Baye’s theorem describes how the conditional probability of each of a set of possible causes for a given observed outcome can be computed from knowledge of the probability of each cause and the conditional probability of the outcome of each cause.
In simple words to get understanding, the probability of an event that occurs after one event occurs can be calculated through this.
In the given problem, we are asked to calculate the probability that the second spring drawn is faulty. So, the spring that we draw first can be anyone. It can be either a good or faulty one.
Let us note down the Baye’s theorem below
$P\left( B \right) = P\left( {B/A} \right)P\left( A \right) + P\left( {B/A'} \right)P\left( {A'} \right)$
In which $P\left( B \right)$ is the probability of the second event and $P\left( A \right)$ is the probability of the first event.
Let $P\left( B \right)$ be the probability of the second spring chosen is faulty and $P\left( A \right)$ is the probability of the first spring chosen is faulty.
Total number of springs in the box $= 8$
Number of faulty springs $= 3$
So, the probability of event $A$ is $\left( {\dfrac{3}{8}} \right)$ and $B$ is $\left( {\dfrac{2}{7}} \right)$
Because there are three faulty springs out of eight and when one is chosen there will be only two faulty springs out of seven.
We know that $P\left( {A} \right) + P\left( {A’} \right) = 1$
So for the event that the first drawn spring is good is $P\left( {A'} \right) = 1-\dfrac{3}{8} = \dfrac{5}{8}$.
Now let’s note down the theorem’s formula once again,
$\Rightarrow P\left( B \right) = P\left( {B/A} \right)P\left( A \right) + P\left( {B/A'} \right)P\left( {A'} \right)$
$P\left( {B/A} \right) = \dfrac{2}{7}$ because there will be two faulty spring out of seven springs after first picking,
$P\left( A \right) = \dfrac{3}{8}$ because there are three faulty springs out of eight springs initially,
$P\left( {B/A'} \right) = \dfrac{3}{7}$ because there will be three faulty springs out of seven if the first choice is a good spring.
$P\left( {A'} \right) = \dfrac{5}{8}$ because there are five good springs out of eight springs initially.
Substituting these values in the Baye’s theorem, we can get the desired value
$\Rightarrow P\left( B \right) = \dfrac{2}{7} \times \dfrac{3}{8} + \dfrac{3}{7} \times \dfrac{5}{8}$
$= \dfrac{6}{{56}} + \dfrac{{15}}{{56}} \\ = \dfrac{{21}}{{56}} \\ = \dfrac{3}{8} \\$
Hence, the probability of the second drawn spring would be faulty is $\dfrac{3}{8}$.

So, the correct answer is “Option C”.

Note:When you see that there are two events like in the problem that we have solved now, it's most probably related to the Bayes theorem but not for the sure for always. You can simply understand this event happening even if you are not familiar with the Bayes theorem.