Question

The content of urn 1, 2 and 3 are as follows:
Urn 1: 1 White, 2 black and 3 red balls
Urn 2: 2 White, 1 black and 1 red balls
Urn 3: 4 White, 5 black and 3 red balls
One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from urn 1, 2 and 3.

Answer 1

Hint: We have to find the probability of getting a white and red ball. We have to use the formulas of probability to solve this question. It is to be noted that all the balls are taken from one urn only.

Complete Step by Step Solution:
According to the question we have to find the probability of getting a white and red ball
So let$E1$,$E2$ and $E3$ be the events of selection of urn 1, urn 2 and urn 3
Now, let A be the event of getting a white and a red ball
So now, we have to find the probability of $E1$,$E2$, $E3$ and A
$P(x)$ denotes the probability of happening of x
So,
$ \Rightarrow P(E1) = \dfrac{1}{3}$
$ \Rightarrow P(E2) = \dfrac{1}{3}$
$ \Rightarrow P(E3) = \dfrac{1}{3}$
Now,
$ \Rightarrow P(\dfrac{A}{{E1}}) = \dfrac{3}{{15}} = \dfrac{1}{5}$
$ \Rightarrow P(\dfrac{A}{{E2}}) = \dfrac{2}{6} = \dfrac{1}{3}$
$ \Rightarrow P(\dfrac{A}{{E3}}) = \dfrac{{12}}{{66}} = \dfrac{2}{{11}}$
Now, we have to use Baye’s theorem
So the probability of getting a white and a red ball when urn 1 is chosen is $ \Rightarrow P(\dfrac{{E1}}{A}) = \dfrac{{P(E1)P(\dfrac{A}{{E1}})}}{{P(E1)P(\dfrac{A}{{E1}}) + P(E2)P(\dfrac{A}{{E2}}) + P(E3)P(\dfrac{A}{{E3}})}}$
$ \Rightarrow P(\dfrac{{E1}}{A}) = \dfrac{{\dfrac{1}{3} \times \dfrac{1}{5}}}{{\dfrac{1}{3} \times \dfrac{1}{5} + \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{{11}}}}$
$ \Rightarrow P(\dfrac{{E1}}{A}) = \dfrac{{33}}{{118}}$
Now, the probability of getting a white and a red ball when urn 2 is chosen is
$ \Rightarrow P(\dfrac{{E2}}{A}) = \dfrac{{P(E2)P(\dfrac{A}{{E2}})}}{{P(E1)P(\dfrac{A}{{E1}}) + P(E2)P(\dfrac{A}{{E2}}) + P(E3)P(\dfrac{A}{{E3}})}}$
$ \Rightarrow P(\dfrac{{E2}}{A}) = \dfrac{{\dfrac{1}{3} \times \dfrac{1}{3}}}{{\dfrac{1}{3} \times \dfrac{1}{5} + \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{{11}}}}$
$ \Rightarrow P(\dfrac{{E2}}{A}) = \dfrac{{55}}{{118}}$
Now, the probability of getting a white and a red ball when urn 3 is chosen is
$ \Rightarrow P(\dfrac{{E3}}{A}) = \dfrac{{P(E3)P(\dfrac{A}{{E3}})}}{{P(E1)P(\dfrac{A}{{E1}}) + P(E2)P(\dfrac{A}{{E2}}) + P(E3)P(\dfrac{A}{{E3}})}}$
$ \Rightarrow P(\dfrac{{E3}}{A}) = \dfrac{{\dfrac{1}{3} \times \dfrac{2}{{11}}}}{{\dfrac{1}{3} \times \dfrac{1}{5} + \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{{11}}}}$
$ \Rightarrow P(\dfrac{{E3}}{A}) = \dfrac{{30}}{{118}}$

Hence the probability of urn one is $ \Rightarrow P(\dfrac{{E1}}{A}) = \dfrac{{33}}{{118}}$
The probability of urn two is $ \Rightarrow P(\dfrac{{E2}}{A}) = \dfrac{{55}}{{118}}$
The probability of urn three is $ \Rightarrow P(\dfrac{{E3}}{A}) = \dfrac{{30}}{{118}}$
This is our answer.

Note:
We have used the Bayes theorem. Essentially, the Bayes' theorem describes the probability Total Probability Rule (also known as the law of total probability) is a fundamental rule in statistics relating to conditional and marginal events based on prior knowledge of the conditions that might be relevant to the event.