Question

In a shop X, 30 tin pure ghee and 40 tin adulterated ghee are kept for sale while in shop Y, 50 tin pure ghee and 60 tin adulterated ghee are there. One tin of ghee is purchased from one of the shops randomly and it is found to be adulterated. Find the probability that it is purchased from shop Y.

Answer 1

Hint:First calculate the probability of purchasing ghee from shop X and shop Y. The probability of purchasing ghee from shop X, $P\left( X \right)$ will be $\dfrac{1}{2}$ and from shop Y, $P\left( Y \right)$ will be $\dfrac{1}{2}$ . In a shop X, 30 tin pure ghee and 40 tin adulterated ghee are there. Now, calculate $P\left( A|X \right)$ , the probability of shop X containing adulterated ghee. After calculation we get,
$P\left( A|X \right)=\dfrac{40}{70}$ . In a shop Y, 50 tin pure ghee and 60 tin adulterated ghee are there. Now, calculate $P\left( A|Y \right)$ , the probability of shop Y containing adulterated ghee. After calculation we get, $P\left( A|Y \right)=\dfrac{60}{110}$ . Then, calculate the total probability of adulterated ghee using the formula, $P\left( A \right)=P\left( X \right).P\left( A|X \right)+P\left( Y \right).P\left( A|Y \right)$ . Now, we have to find the probability that the adulterated ghee is purchased from shop Y. It means we have to find the probability, $P\left( Y|A \right)$ . Using Bayes theorem, we can write it as $P\left( Y|A \right)=\dfrac{P\left( A|Y \right).P\left( Y \right)}{P\left( A \right)}$ . Now, put the values of $P\left( X \right)$ , $P\left( Y \right)$ , $P\left( A|X \right)$ , and $P\left( A|Y \right)$ in the Bayes theorem formula, and solve it further.

Complete step-by-step answer:
According to the question, it is given that we have two shops which are shop X and shop Y.
The total number of shops = 2.
The probability of purchasing ghee from shop X, $P\left( X \right)$ = $\dfrac{1}{2}$ ……………….(1)
The probability of purchasing ghee from shop Y, $P\left( Y \right)$ = $\dfrac{1}{2}$ …………………(2)
It is given that in shop X we have 30 tin pure ghee and 40 tin adulterated ghee.
The number of tins of pure ghee = 30 tins.
The number of tins of adulterated ghee = 40 tins.
The total number of tins of ghee = 30 tins + 40 tins = 70 tins.
In shop X, the probability of adulterated ghee, $P\left( A|X \right)$ = $\dfrac{40}{70}=\dfrac{4}{7}$ …………..(3)
It is given that in shop Y we have 50 tin pure ghee and 60 tin adulterated ghee.
The number of tins of pure ghee = 50 tins.
The number of tins of adulterated ghee = 60 tins.
The total number of tins of ghee = 50 tins + 60 tins = 110 tins.
In shop Y, the probability of adulterated ghee, $P\left( A|Y \right)$ = $\dfrac{60}{110}=\dfrac{6}{11}$ ……………(4)
The tin of adulterated ghee can be either from shop X or shop Y.
The total probability of adulterated ghee,
$P\left( A \right)=P\left( X \right).P\left( A|X \right)+P\left( Y \right).P\left( A|Y \right)$
Putting the values of $P\left( X \right)$ , $P\left( Y \right)$ , $P\left( A|X \right)$ , and $P\left( A|Y \right)$ in the above equation, we get,
$P\left( A \right)=P\left( X \right).P\left( A|X \right)+P\left( Y \right).P\left( A|Y \right)$
$\Rightarrow P\left( A \right)=\dfrac{1}{2}\left( \dfrac{4}{7} \right)+\dfrac{1}{2}\left( \dfrac{6}{11} \right)$ ………………(5)
We know the Bayes theorem, $P\left( A|B \right)=\dfrac{P\left( B|A \right).P\left( A \right)}{P\left( B \right)}$ .
Now, we have to find the probability that the adulterated ghee is purchased from shop Y. It means we have to find the probability, $P\left( Y|A \right)$ .
Now, using Bayes theorem, we can write $P\left( Y|A \right)$ as,
$P\left( Y|A \right)=\dfrac{P\left( A|Y \right).P\left( Y \right)}{P\left( A \right)}$ ……………………(6)
Now, from equation (1), equation (2), equation (4), and equation (5), we have
$P\left( Y \right)$ = $\dfrac{1}{2}$ , and $P\left( A|Y \right)$ = $\dfrac{6}{11}$ ,and $P\left( A \right)=\dfrac{1}{2}\left( \dfrac{4}{7} \right)+\dfrac{1}{2}\left( \dfrac{6}{11} \right)$ .
Puuting these values in equation (6), we get

& P\left( Y|A \right)=\dfrac{P\left( A|Y \right).P\left( Y \right)}{P\left( A \right)} \\\ & \Rightarrow P\left( Y|A \right)=\dfrac{\dfrac{6}{11}\times \dfrac{1}{2}}{\dfrac{1}{2}\left( \dfrac{4}{7} \right)+\dfrac{1}{2}\left( \dfrac{6}{11} \right)} \\\ & \Rightarrow P\left( Y|A \right)=\dfrac{\dfrac{6}{22}}{\dfrac{44+42}{2\times 11\times 7}} \\\ & \Rightarrow P\left( Y|A \right)=\dfrac{6\times 77\times 2}{22\times 86} \\\ & \Rightarrow P\left( Y|A \right)=\dfrac{7\times 3}{43} \\\ & \Rightarrow P\left( Y|A \right)=\dfrac{21}{43} \\\ \end{aligned}$$ Hence, the probability that adulterated ghee is purchased from shop Y. Note: In this question, one might get the probability $$P\left( A|Y \right)$$ and conclude it as an answer which is wrong. The probability $$P\left( A|Y \right)$$ is the probability of shop Y containing adulterated ghee. But we have to calculate the probability that adulterated ghee is purchased from shop Y, $$P\left( Y|A \right)$$ . Therefore, the probabilities $$P\left( A|Y \right)$$ and $$P\left( Y|A \right)$$ are two different probabilities.