Urn A contains five white balls and seven black balls. Urn B contains three white balls and four black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was black given that the second ball drawn was black? (Round your answer to three decimal places.)

Answer:0.636Step-by-step explanation:Let Aw, Ab, Bw, Bb the following events: Aw = the ball drawn from urn A is white Ab = the ball drawn from urn A is black Bw = the ball drawn from urn B is white Bb = the ball drawn from urn B is black the probability that the transferred ball was black given that the second ball drawn was black is P(Ab | Bb) By the Bayes' Theorem [tex]\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}[/tex] Working out the calculations [tex]\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}=\\\frac{35/96}{35/96+20/96}=35/55=7/11=0.636[/tex] rounded to three decimals