Urn A contains 2 red and 4 white balls, and urn B contains 1 red and 1 white ball. A ball is randomly chosen from urn A and put into urn B, and a ball is then chosen from urn B. What is the conditional probability that the transferred ball was white given that a white ball is selected from urn B?

Answer: 0.7991Step-by-step explanation:Given : Urn A contains 2 red and 4 white balls, and urn B contains 1 red and 1 white ball.Let [tex]E_1[/tex] be the event that the transferred ball was white.and [tex]E_2[/tex] be the event that the ball drawn from urn B was white.Then by Bayes theorem, the conditional probability that the transferred ball was white given that a white ball is selected from urn B will be :-[tex]P(E_1|E_2)=\dfrac{P(E_1\cap E_2)}{P(E_2)}[/tex]Where, [tex]P(E_1\cap E_2)=P(E_2|E_1) P(E_1)=\dfrac{2}{3}\times\dfrac{4}{6}=0.4444[/tex] [ by conditional probability formula][tex]P(E_2)=P(E_2|E_1)\times P(E_1)+P(E_2|E_1^c)\times P(E_1^c)[/tex] [By law of total probability]i.e. [tex]P(E_2)=\dfrac{2}{3}\times \dfrac{4}{6}+\dfrac{1}{3}\times \dfrac{2}{6}=0.5556[/tex]Now, [tex]P(E_1|E_2)=\dfrac{0.444}{0.5556}=0.79913606911\approx0.7991[/tex]Hence, the required probability =0.7991