Question

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 1

Answer:

0.636

Step-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

$\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}$

Working out the calculations

$\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}=\\\frac{35/96}{35/96+20/96}=35/55=7/11=0.636$

rounded to three decimals