State Bayes theorem. Three urns A, B, C contains white, red and green balls as given below. Two balls are drawn from an urn chosen at random. These are found to be one white and one Green. Find the probability that the balls drawn are from urn C.
Urn A
Urn C
White
1
2
4
Red
2
1
5
Green
3
Urn B
1
3
Bayes' Theorem states that the probability of an event A given event B is equal to the probability of event B given event A, multiplied by the probability of event A, divided by the probability of event B.
Let A be the event that the two balls drawn are from urn C, and B be the event that one ball is white and one ball is green.
We are given that urn C contains 2 white balls and 3 green balls. The probability of drawing a white ball from urn C is 2/5 and the probability of drawing a green ball from urn C is 3/5.
The total probability of drawing one white and one green ball from any urn can be calculated as follows:
P(B) = P(WW) + P(WG) + P(GW) + P(GG)
= (1/5) * (2/5) + (1/5) * (3/5) + (3/5) * (1/5) + (3/5) * (2/5)
= 2/25 + 3/25 + 3/25 + 6/25
= 14/25
The probability of drawing one white and one green ball from urn C can be calculated as follows:
P(B|A) = P(WG|C) + P(GW|C)
= (2/5) * (3/5) + (3/5) * (2/5)
= 6/25 + 6/25
= 12/25
Now, applying Bayes' Theorem:
P(A|B) = P(B|A) * P(A) / P(B)
= (12/25) * (1/3) / (14/25)
= 12/42
= 2/7
Therefore, the probability that the balls drawn are from urn C is 2/7.