Urn 1 and Urn 2

Question

Urn 1 contains two white balls and one black ball, while urn 2 contains one white ball and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball is white?

Answer 1

Urn 1: 2 white 1 black balls

Urn 2 1 white 5 black

1 ball is randomly chosen and moved from Urn 1 to Urn 2

then a ball is randomly chosen from Urn 2

what are the odds it is white?? 4/5= 80%= every other posted answer

but real answer is seemingly 5/3/7 = 5/21 chance = 23.80953...%

7 balls in Urn 2, 5 black 1 white and 1 2/3 chance of white

that's 1 2/3 chance of white out of 7 = 5/21

but that misreads wth question. question was if the ball chosen was white what are the odds the transferred ball was white and that is 4/5

use Bayes Theorem

P(a)P(b/a) = P(b)P(a/b) where a= white= 5/21 b=white transferred from Urn 1 to 2 = 2/3

a/b = white given white transferred= 2/7, b/a = white transferred given white randomly selected

find P(b/a) = (2/7)(2/3)/(5/21) = (4/21)/(5/21) = 4/5 = 80% chance

Answer 2

W_T​: The transferred ball is white.

W₂: A white ball is drawn from urn 2.

Urn 1 has 2 white balls and 1 black ball, so:

P(W_T) = 2/3

The probability of transferring a black ball from urn 1:

P(W_T^c) = 1 − P(W_T) = 1/3

If a white ball is transferred, urn 2 will have 2 white balls and 5 black balls. The probability of drawing a white ball is: P(W₂∣W_T) = 2/7

If a black ball is transferred, urn 2 will have 1 white ball and 6 black balls. The probability of drawing a white ball is: P(W₂∣W_T^c) = 1/7

The total probability of drawing a white ball from urn 2 using the law of total probability:

P(W₂) = P(W_T) ⋅ P(W₂∣W_T) + P(W_T^c) ⋅ P(W₂∣W_T^c) = 5/21

Bayes' theorem states:

P(W_T∣W₂) = (P(W_T) ⋅ P(W₂∣W_T)) / P(W₂) = 4/5

The probability that the transferred ball is white, given that a white ball was drawn from urn 2, is:

4/5

Hope this was helpful.

Answer 3

This is Conditional probability

Case #1: urn1 - ball was black

Now, urn 2 has 1 w, 6 b => P(urn 1 is black) * P(urn 2 is W) = 1/3 * 1/7 = 1/21

Case #2: urn1 - ball was white

Now, urn 2 has 2 w, 5 b => P(urn 1 is white) * P(urn 2 is W) = 2/3 * 2/7 = 4/21

NOW, by case 1 and 2, P(the transferred ball is white l urn 2 is white)

the transferred ball is white = urn 1 ball was white

P(the transferred ball is white l urn 2 is white) = 4/21 / (1/21 + 4/21) = 4/5

Answer 4

As soon as I saw the word "urn", I'd turn to my scrap paper (or MS Paint/Wyzant Whiteboard if I was on a laptop). I would draw two urns, label them, add three balls to urn 1 and shade two of them black, add six balls to urn 2 and shade five of them black. I'd then draw and arrow from the top of urn 1 to urn 2, and finally an arrow pointing up out of the top of urn 2. I'd then label the arrow from urn 1 to urn 2 "P1" and the up-arrow "P2".

I would then pause and prompt the student to identify the tool set the question is signalling them to use.

After that conversation (or if they immediately spot "Bayes" or "conditional") I'd also splash down "Bayes/Conditional" and the Bayes' theorem at the top of the page, since this is a question concerning the conditional probability of a prior event given knowledge of a subsequent outcome.

In the specific context of the problem, we have two events: the pick from the first urn (P1) and the pick from the second urn (P2). We're asked for the probability that P1 = White given (or conditioned on) that P2 = White.

In the syntax of Bayes' theorem, we would write this probability as such:

P (P1 = White | P2 = White) = P (P2 = White | P1 = White) * [P(P1 = White) / P(P2 = White)]

If the student recalled the definitions for marginal and intersection probabilities using conditional probabilities, they may note that the value on every variable on the right-hand side of the equation above is determined by the information in the initial question. Depending on the student's knowledge and/or their confidence, I may prompt them to think about what information the problem statement has given us, and to compare that information to the right hand side probabilities. I might also break the right hand side into its component parts. I may also remind them of any reference material or handouts we've been using.

Since this is a post, I'll give that information here. We need the following definition:

P(A) = P(A & B1) + P(A & B2)

= P(A|B1)*P(B1) + P(A|B2)*P(B2)

At this point, we can fill in the right-hand side of Bayes' formula with information from the problem statement.

P(A) = P(P1 = White) = 2/3

P(B|A) = P(P2 = White | P1 = White) = 2/7

P(B) = P(P2 = White) = P(P1 = White & P2 = White) + P(P1 = White & P2 = B) =

P(P2 = White | P1 = White)*P(P1 = White) + P(P2 = White | P1 = Black)*P(P1 = Black) =

(2/7*2/3) + (1/7*1/3) = 5/21

Thus,

P(P1 = W | P2 = W) = (2/7) * (2/3 / 5/21) = (2/7) * (14/5) = (28/35) = 4/5.

Of course, if the student prefers to work with decimals, I would restructure this response throughout to meet their preference, ultimately leading to an answer of 0.8 or 80%.

After arriving at this solution, I would ask them to think about if this solution makes sense. I'd be looking for some reasoning about the 'base' probability (or 'prior' if they're using the class's language) of P1 being white to be 2/3 or 66.66...%, and the additional knowledge that a white ball was drawn in P2 should increase the probability that P1 was white as well. I'd be looking for the keyword "dependent" in the explanation.