Use Bayes s Rule to Compute Probabilities
-Suppose that there are two buckets. Bucket 1 contains 3 tennis balls and 5 ping -pong balls. Bucket 2 contains 4 tennis balls, 3 ping-pong balls, and 5 baseballs. An unfair coin will decide from which bucket we draw. Heads implies we draw from Bucket 1 and tails implies we draw from Bucket 2. The probability of a heads is $\frac { 1 } { 3 }$ and the probability of a tails is $\frac { 2 } { 3 }$. Given that a ping-pong ball was selected what is the probability that it came from Bucket 2?
A) $0.444$
B) $0.555$
C) $0.625$
D) $0.250$

Question

Quizplus · Accepted Answer

Let E be the event that a ping-pong ball was selected and B2 be the event that the ball came from Bucket 2. Then, by Bayes' Rule,$$P(B2|E) = \frac{P(E|B2)P(B2)}{P(E)}$$We know that P(E|B2) = 3/12, since there are 3 ping-pong balls in Bucket 2 and 12 balls total in Bucket 2. We also know that P(B2) = 2/3, since the probability of drawing from Bucket 2 is 2/3. Finally, we need to find P(E), the probability of selecting a ping-pong ball. This is given by$$P(E) = P(E|B1)P(B1) + P(E|B2)P(B2)$$where B1 is the event that the ball came from Bucket 1. We know that P(E|B1) = 5/8, since there are 5 ping-pong balls in Bucket 1 and 8 balls total in Bucket 1. We also know that P(B1) = 1/3, since the probability of drawing from Bucket 1 is 1/3. Plugging these values into the equation, we get$$P(E) = (5/8)(1/3) + (3/12)(2/3) = 5/24 + 1/6 = 11/24$$Now we can plug all of these values into Bayes' Rule to get$$P(B2|E) = \frac{(3/12)(2/3)}{11/24} = \frac{1}{11} \approx 0.091$$Therefore, the probability that the ping-pong ball came from Bucket 2 is approximately 0.091, or 9.1%.

Use Bayes S Rule to Compute Probabilities -Suppose That There 13\frac { 1 } { 3 }31​

Use Bayes S Rule to Compute Probabilities
-Suppose That There $\frac { 1 } { 3 }$