You have 3 bags each containing two marbles.

Bag A contains 2 white marbles ,

Bag B contains 2 black marbles,

Bag C contains one white marble and one black marble.

You pick a random bag and take out one marble. It is a white marble.

What is the Probability that the remaining marble from the same bag is also white?

My Approach:

A:Probability from choosing marble from bag $A=1/3$

B:Probability from choosing marble from bag $B=1/3$

C:Probability from choosing a marble from bag $C=1/3$

D:Probability of a white marble





Is it Correct?

correct me if i am wrong

  • $\begingroup$ Given that you selected a white marble on the first draw, you could not have selected bag B. $\endgroup$ – N. F. Taussig Aug 25 '15 at 1:36
  • $\begingroup$ @N.F Taussing Yes $\endgroup$ – Jack Aug 25 '15 at 1:42
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    $\begingroup$ You could do a formal conditional probability calculation. Let $W$ be the event first marble picked was white and $A$ be the event the marble came from $A$. We want $\Pr(A|W)$, which is $\frac{\Pr(A\cap W)}{\Pr(W)}$. Each of the probabilities on the right is not hard to compute. $\endgroup$ – André Nicolas Aug 25 '15 at 1:46
  • $\begingroup$ @AndréNicolas Sir What is your Ans My Ans is 1/3 according to your logic.Please Explain in detail here too :P . $\endgroup$ – Jack Aug 25 '15 at 1:48
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    $\begingroup$ $\Pr(A\cap W)=1/3$, $\Pr(W)=1/3+1/6=1/2$, ratio simplifies to $2/3$. $\endgroup$ – André Nicolas Aug 25 '15 at 1:51

There are three white marbles.   You have selected one of these.   You want the probability that the one you selected comes from bag A, which held two of the three.

Or if you prefer to do it the hard way, use Bayes' Rule:

$$\mathsf P(A\mid W) = \frac{\mathsf P(A)\mathsf P(W\mid A)}{\mathsf P(W)} = \frac{\tfrac 1 3\cdot\tfrac 2 2}{\tfrac 3 6} = \frac 2 3 $$

  • $\begingroup$ @Grham Kemp Not understood please explain in detail. $\endgroup$ – Jack Aug 25 '15 at 1:44
  • $\begingroup$ I think he means that the only way you can select a white marble and then another white would be if you selected bag A so the probability is $1/3$, the probability of selecting bag A @Jack $\endgroup$ – Blake Aug 25 '15 at 1:46
  • $\begingroup$ @Black No. Jack cannot have selected bag B, and bag A contained twice as many white marbles as bag C (all equally likely to have been chosen). $\endgroup$ – Graham Kemp Aug 25 '15 at 1:52

2/3. There are 3 possibilities, 2 of which are from bag A and 1 from bag C.

There are 3 marbles you might have picked :

AW1, AW2, CW

2/3 of these (the first 2) are from a bag where the other is white. Make sense?

  • $\begingroup$ Why my ans is wrong i haven't understood. $\endgroup$ – Jack Aug 25 '15 at 1:46
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    $\begingroup$ I edited, please respond if it's now understood. $\endgroup$ – JoeTaxpayer Aug 25 '15 at 1:51
  • $\begingroup$ yes it is understood :) $\endgroup$ – Jack Aug 25 '15 at 2:05

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