Urn A contains four white balls and seven black balls. Urn B contains six white balls and three 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 white given that the second ball drawn was white? (Round your answer to three decimal places.)
Conceptually this is the same category of problems as the previous boomers problem.
Let events
w=a white ball was transferred
b=a black ball was transferred
now concentrate on urn B.
W=a white ball was drawn
B=a black ball was drawn
then we know that
probability of transferring a white ball
P(w)=4/11
and
P(b)=7/11.
If a white ball was transferred, then probability of drawing a white ball from urn B
P(W)=7/10, or
P(W|w)=7/10
and similarly,
P(W|b)=6/10
Applying the law of total probability
P(W)=P(W|w)*P(w)+P(W|b)*P(b)
=(7/10)(4/11)+(6/10)*(7/11)
=7/11
(appreciate the advantages of working with fractions in probability calculations)
Applying Bayes theorem
P(w|W)=P(W|w)*P(w)/P(W)
=(7/10)*(4/11)/(7/11)
=(28/110)/(7/11)
=2/5
To calculate the probability that the transferred ball was white given that the second ball drawn was white, we need to use the concept of conditional probability.
Let's solve this step-by-step:
Step 1: Determine the probability of drawing a white ball from Urn A and transferring it to Urn B:
The probability of drawing a white ball from Urn A is 4/(4+7) = 4/11.
Step 2: Determine the probability of drawing a white ball from Urn B:
After transferring the ball from Urn A, Urn B contains a total of (6+1 white balls from Urn A) + 3 = 10 balls. The probability of drawing a white ball from Urn B is 6/10 = 3/5.
Step 3: Calculate the conditional probability using Bayes' Theorem:
The conditional probability can be calculated using Bayes' Theorem:
P(white ball transferred from Urn A | second ball drawn from Urn B is white) = P(white ball transferred from Urn A and second ball drawn from Urn B is white) / P(second ball drawn from Urn B is white)
P(white ball transferred from Urn A and second ball drawn from Urn B is white) = (4/11) * (6/10) = 24/110
P(second ball drawn from Urn B is white) = (6/10) * (1/2) + (4/10) * (4/7) = 23/35
P(white ball transferred from Urn A | second ball drawn from Urn B is white) = (24/110) / (23/35)
Finally, calculating the result:
P(white ball transferred from Urn A | second ball drawn from Urn B is white) ≈ (24/110) / (23/35) ≈ 0.876 (rounded to three decimal places)
Therefore, the probability that the transferred ball was white given that the second ball drawn from Urn B was white is approximately 0.876.
To find the probability that the transferred ball was white given that the second ball drawn was white, we need to use conditional probability.
Let's break down the problem step by step:
Step 1: Calculate the probability of drawing a white ball from Urn A.
Urn A contains 4 white balls and 7 black balls, so the probability of drawing a white ball from Urn A is 4/11.
Step 2: Calculate the probability of drawing a white ball from Urn B, given that the transferred ball was white.
After transferring the ball from Urn A to Urn B, Urn B will contain a total of 5 white balls (original 6 white balls - 1 transferred) and 3 black balls. Therefore, the probability of drawing a white ball from Urn B, given that the transferred ball was white, is 5/8.
Step 3: Calculate the probability that the transferred ball was white given that the second ball drawn was white.
This is where we use conditional probability. The probability is calculated as the product of the probabilities from Step 1 and Step 2 divided by the probability of drawing a white ball from Urn B. So the probability can be calculated as follows:
(4/11) * (5/8) / (5/8) = 4/11
Therefore, the probability that the transferred ball was white given that the second ball drawn was white is 4/11, rounded to three decimal places.