Consider urn 1: 2 white balls , 3 black balls
Urn 2 : 4 white balls, 6 black balls
One ball is randomly transfer from first to second urn, then one ball is drawn from urn 2. The probability that drawn ball is white ??????
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
In the above case, there are 2 possible outcomes:-
Outcome 1 – 1 White ball was transferred from Urn 1 and then White ball was drawn from Urn 2.
Outcome 2 – 1 Black ball was transferred from Urn 1 and then the White ball was drawn from Urn 2.
We need to calculate the probability of the above 2 possibilities and add them to get the total probability of drawing a white ball.
P1 – The probability of transferring 1 white ball to Urn 2 is 2/5.
P2 – The probability of transfereing 1 black ball in Urn 2 is 3/5.
X1 – If P1 happens, the no. of white balls in the Urn 2 will increase to 5 and the individual probability of drawing 1 white ball will be 5/11.
Therefore, the probability of choosing 1 white ball, given that 1 white ball was transferred from Urn 1 to Urn 2 is (2/5 * 5/11) = 2/11 (.1819 or 18.19% approx)
X2 – If P2 happens, the no. of black balls in the Urn 2 will increase to 7 and the individual probability of drawing 1 white ball will be 4/11.
Therefore, the probability of choosing 1 white ball, given that 1 black ball was transferred from Urn 1 to Urn 2 is (3/5 * 4/11) = 12/55 (.2182 or 21.82% approx)
Now the combined probability of drawing a white ball will be X1 + X2
= 2/11 + 12/55
= 22/55 (.40 or 40% approx)
Hope this helps.