Question

A bag consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. A trader A will accept only the shirts which are good, but the trader B will not accept the shirts which have major defects. One shirt is drawn at random. What is the probability that it is acceptable by (i) A (ii) B?
A. $\left( i \right)\dfrac{21}{25}\text{ }\left( ii \right)\dfrac{23}{25}$
B. $\left( i \right)\dfrac{22}{25}\text{ }\left( ii \right)\dfrac{24}{25}$
C. $\left( i \right)\dfrac{11}{25}\text{ }\left( ii \right)\dfrac{13}{25}$
D. None of these

Answer 1

Hint: The total number of shirts are given as 100 and the number of good shirts, shirts with minor defects and with major defects are also given. So, we will first find the probability for each type of shirt by using the formula, $\dfrac{\text{favourable outcomes of event}}{\text{total number of outcomes}}$. We will consider $P\left( G \right)$ as the probability of drawing good shirts, $P\left( M \right)$ as probability of drawing minor defects shirts and $P\left( D \right)$ as the probability of drawing major defects shirts. Now, it is given that A will accept only good shirts, so $P\left( A \right)=P\left( G \right)$ and B will not accept shirts with major defects, which means he will accept the good shirts and the minor defects shirts, so, $P\left( B \right)=P\left( G \right)+P\left( M \right)$. We can find these and get the answer.

Complete step-by-step answer:
We are given that the total number of shirts are 100, out of which 88 are good, 8 have minor defects and 4 have major defects. It is also given that trader A will accept only the shirts which are good, but the trader B will not accept the shirts which have major defects and we have to find the probability that a shirt drawn at random is accepted by A and B. So, we will first find the probability of drawing each type of shirts by using the formula, $\dfrac{\text{favourable outcomes of event}}{\text{total number of outcomes}}$.
So, we will find the probability of drawing a good shirt. There are 88 good shirts and the total number of shirts is 100, so we get,
$\begin{align}
  & P\left( G \right)=\dfrac{88}{100} \\
 & \Rightarrow P\left( G \right)=\dfrac{22}{25}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Now, we will find the probability of drawing a shirt with minor defects, $P\left( M \right)$. There are 8 such shirts. So, we get,
$\begin{align}
  & P\left( M \right)=\dfrac{8}{100} \\
 & \Rightarrow P\left( M \right)=\dfrac{2}{25}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
We will now find the probability of drawing a shirt with major defects, $P\left( D \right)$, which are 4 in number. So, we get,
$\begin{align}
  & P\left( D \right)=\dfrac{4}{100} \\
 & \Rightarrow P\left( D \right)=\dfrac{1}{25}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
Now we will find the probability according to the conditions given in the question.
(i) Shirt acceptable by trader A.
Now, we have been given that trader A will accept only good shirts, so the probability of A accepting good shirts, $P\left( A \right)$ will be,
$P\left( A \right)=P\left( G \right)$
We know the value of $P\left( G \right)$ from equation (i) so we can write the above as,
$P\left( A \right)=\dfrac{22}{25}$
Hence, we can say that the probability of trader A accepting a shirt is $\dfrac{22}{25}$.
(ii) Shirt acceptable by trader B.
Now, we are given that trader B will not accept shirts with major defects, which means he will accept shirts that are good and have minor defects. So, the probability of trader B accepting the shirts $P\left( B \right)$ will be given as,
$P\left( B \right)=P\left( G \right)+P\left( M \right)$
We have the values of $P\left( G \right)$ and $P\left( M \right)$ from equations (i) and (ii), so we can substitute them. So, we get,
$\begin{align}
  & P\left( B \right)=\dfrac{22}{25}+\dfrac{2}{25} \\
 & \Rightarrow P\left( B \right)=\dfrac{24}{25} \\
\end{align}$
Hence, the probability of trader B accepting a shirt is $\dfrac{24}{25}$.
Therefore, the correct answer is option B.

Note: We can solve the second part of the question using an alternate method too. We have been given that trader B will not accept shirts that have major defects. Now, we know that the probability of total shirts = 1. So, probability that B will accept the shirt will be = 1 - probability (of B not accepting a shirt), which is,
$P\left( B \right)=1-P\left( D \right)$
We have the value of $P\left( D \right)$ from equation (iii), so we can write as,
$\begin{align}
  & P\left( B \right)=1-\dfrac{1}{25} \\
 & \Rightarrow P\left( B \right)=\dfrac{25-1}{25} \\
 & \Rightarrow P\left( B \right)=\dfrac{24}{25} \\
\end{align}$
So, we get the probability of trader B accepting a shirt as $\dfrac{24}{25}$.