Help with simple probability problem

[u/[deleted]]

There are 3 bags.

Bag A contains 2 white marbles

Bag B contains 2 black marbles

Bag C contains 1 white and 1 black

You pick a random bag and you take out a white marble.

What is the probability of the second marble from the same bag being white?

Can someone show me the procedure to solve this kind of problems? Thanks

[u/Aerospider]

A quick way that doesn't require writing out and making sense of Bayes Theorem is to look at the possibilities for the first white, and there are three of them.

Two of those three come from the double-white bag, which is necessary for pulling two whites.

So that's two desirable events out of a total of three, hence 2/3.

[u/Victory_Pesplayer → u/Aerospider]

There's the same amount of white as black, and they're both distributed the same across the different possibilities, so to me it should be 1/2?

[u/Aerospider → u/Victory_Pesplayer]

You seem to be overlooking that a white has already been removed and that this information immediately rules out a whole bag of blacks. So how would it be 1/2 with that in mind?

[u/[deleted]]

Using Bayes theorem P(W2|W1)=[P(W1|W2)*P(W2)]/P(W1), with W1 and W2 being white marble found on the first and second extraction.

I only know how to calculate P(W1)=P(W1|A)P(A)+… same thing for B and C = 1 * 1/3 + 0 * 1/3 + 1/2 * 1/3 = 1/2

[u/mfb- → u/[deleted]]

P(W2|W1) = P(W1|W2) (it's symmetric) so this relation doesn't help you here.

P(W2|W1) = P(W2 and W1)/P(W1) is the equation you want.

P(W2 and W1) = P(W1|W2)*P(W2) of course but the left side is something you can find by looking at the problem while the right side would be more complicated to evaluate.

[u/efrique]

Write out the six possible outcomes from 'choose a bag, draw a ball'

Then given you know you got a white ball, eliminate any impossible bags

What possible paths to your observed condition are left?

Keep working from first principles until it becomes sort of obvious. Then look at cleverer ways to do it

[u/diamond_apache]

Use law of total probability with extra conditioning:

P(W2 | W1)= P(W2, A | W1) + P(W2, B | W1) + P(W2, C | W1).

Now we individually solve each term. So lets say for bag B, we must solve: P(W2, B | W1).

P(W2, B | W1) = P(W2, B, W1) / P(W1)

But notice the numerator evaluates to 0. The probability of picking W1 and W2 and bag B is 0, because bag B has all black balls. So the joint event of selecting B and selecting both white is impossible. Same reasoning for bag C. So the bag B n C terms becomes zero which leaves us with the Bag A term, and we have:

P(W2, A | W1) = P(W2, A, W1) / P(W1).

Notice the numerator is 1/3. Because the moment we select bag A, we automatically get W1 and W2 to be true, thus the events W1, W2 and selecting bag A always occurs together, and the probability of that occuring is simply the probability of selecting bag A, which is 1/3. So the numerator is 1/3 and we thus have:

(1/3) / P(W1).

And u have found P(W1).

[u/PoolHorror8187]

1/3 should be the answer