Question

The contents of urns A,B,C are as follows:
Urn A: 1 white,2 black and 3 red balls
Urn B: 2 white,1 black and 1 red balls
Urn C: 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 Urns A,B,C?

Answer 1

Hint: To find the probability that they come from three urns, we first need to calculate the individual probabilities so that they can be drawn from Urn A, B, C respectively. And then we can add the three probabilities to get our required answer.

Complete step by step answer:
According to the question, the total balls in Urn A is 6 balls and the total balls in Urn B is 4 balls and the total number of balls in Urn C are 12 balls.
First we will find the probability that the chosen Urn is A and two balls are drawn randomly are white and red from Urn A:
Which equals,
(Probability that the Urn chosen is A)$ \times $ (the probability that the two balls are drawn are white and red)
=$\dfrac{1}{3} \times [\dfrac{{^1{C_1}{.^3}{C_1}}}{{^6{C_2}}}]$
Now we will find the probability that the chosen Urn is B and two balls drawn randomly are white and red from Urn B:
Which equals,
(Probability that the Urn chosen is B)$ \times $ (probability that the two balls drawn are white and red)
=$\dfrac{1}{3} \times [\dfrac{{^2{C_1}{.^1}{C_1}}}{{^4{C_2}}}]$

 Finally, we will find the probability that the chosen Urn is C and two balls are drawn randomly are white and red from Urn C:
Which equals,
(Probability that the Urn chosen is C)$ \times $ (the probability that the two balls are drawn are white and red)
=$\dfrac{1}{3} \times [\dfrac{{^4{C_1}{.^3}{C_1}}}{{^{12}{C_2}}}]$
Now we can solve the above equations and add it together for the final solution
Which implies,
The probability that one white and one red ball comes from Urns A, B, C when an Urn is selected
randomly will be

=$\dfrac{1}{3} \times [\dfrac{{^1{C_1}{.^3}{C_1}}}{{^6{C_2}}}]$+$\dfrac{1}{3} \times [\dfrac{{^2{C_1}{.^1}{C_1}}}{{^4{C_2}}}]$+$\dfrac{1}{3} \times [\dfrac{{^4{C_1}{.^3}{C_1}}}{{^{12}{C_2}}}]$
=$\dfrac{1}{3}[\dfrac{3}{{15}} + \dfrac{2}{6} + \dfrac{{12}}{{66}}]$
=$\dfrac{1}{3} \times [\dfrac{{66 + 110 + 60}}{{330}}]$
=$\dfrac{{236}}{{990}}$
=\[\dfrac{{118}}{{495}}\]

Note:
 Make sure to read the question clearly while solving problems related to permutations and combinations because there are many chances that you can misunderstand the question. Do not make any calculation mistakes while solving the problem. And check whether you need to multiply or add the result.