What Is the Probability of Drawing a Specific Marble Color Sequence?

AI Thread Summary
The problem involves drawing marbles from an urn containing two white marbles and one black marble without replacement. Given that the second marble drawn is white, the probabilities of the first marble being white or black are debated, with calculations suggesting various probabilities. The correct application of Bayes' theorem indicates that the probability of the first marble being white is 0.6667, while the probability of it being black is 0.333. Suggestions for further understanding include simulating the scenario and using probability trees to visualize outcomes. The discussion emphasizes careful calculation and consideration of the order of events in determining probabilities.
Tajeshwar
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Homework Statement



An Urn contains two white marbles and one black marble. A marble is drawn from the Urn without replacement and put aside without my seeing it. Then a second marble is drawn, and it is white.
What is the probability that the unknown removed marble is white, and what is the probability that it is black?

p(the first marble is white∣the second marble is white)=1.0
p(the first marble is black∣the second marble is white)=0.0

p(the first marble is white∣the second marble is white)=0.6667
p(the first marble is black∣the second marble is white)=0.333

p(the first marble is white∣the second marble is white)=0.3333
p(the first marble is black∣the second marble is white)=0.6667

p (the first marble is white |the second marble is white) =.5
p(the first marble is black | the second marble is white) =.5

Homework Equations

The Attempt at a Solution



My attempt is as follows:

P(first black ("FB") |second white ("SW")) = P(SW|FB)*P(FB)/(P(SW|FB)*P(FB)+P(SW|FW)*P(FW))

Using Bayes theorem.

This gives me = (1/2)*(1/3)/(1/6+1/3) = 0.33

Did a similar logic for P(First White|Second White) and got .66

But I am getting this question wrong when I pick the following answer:

p(the first marble is white∣the second marble is white)=0.6667
p(the first marble is black∣the second marble is white)=0.333
 
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My suggestion is to simulate this problem experimentally. Perhaps with playing cards. Two red and one black, say.

Run through the experiment until you see what is happening. This will give you the numerical answer.

Note: this is actually a well known problem and well worth studying.

PS you can also simulate this simply by imagining running the experiment 12 times say and counting how often the first ball is white when the second ball is white.
 
Last edited:
What number are you using for P(SW|FB)?
 
Also, again, I would use a probability tree here.
 
Tajeshwar said:

Homework Statement



An Urn contains two white marbles and one black marble. A marble is drawn from the Urn without replacement and put aside without my seeing it. Then a second marble is drawn, and it is white.
What is the probability that the unknown removed marble is white, and what is the probability that it is black?

p(the first marble is white∣the second marble is white)=1.0
p(the first marble is black∣the second marble is white)=0.0

p(the first marble is white∣the second marble is white)=0.6667
p(the first marble is black∣the second marble is white)=0.333

p(the first marble is white∣the second marble is white)=0.3333
p(the first marble is black∣the second marble is white)=0.6667

p (the first marble is white |the second marble is white) =.5
p(the first marble is black | the second marble is white) =.5

Homework Equations

The Attempt at a Solution



My attempt is as follows:

P(first black ("FB") |second white ("SW")) = P(SW|FB)*P(FB)/(P(SW|FB)*P(FB)+P(SW|FW)*P(FW))

Using Bayes theorem.

This gives me = (1/2)*(1/3)/(1/6+1/3) = 0.33

Did a similar logic for P(First White|Second White) and got .66

But I am getting this question wrong when I pick the following answer:

p(the first marble is white∣the second marble is white)=0.6667
p(the first marble is black∣the second marble is white)=0.333

Let the relevant events be W1={first is white}, B1= {first is black}, W2 = {second is white}.

Can you compute P(W1 & W2), P(B1 & W2)? If so, can you see how to get P(W2)?

Then, you ought to be able to compute P(W1|W2) and P(B1|W2).

Finally: avoid writing curtailed decimals; better to write 2/3 instead of .66 or .667. You can always round off the final answer, but it is a dangerous practice to round off too much during a series of calculations (although keeping full calculator accuracy throughout should help).
 
After answering post #3 to find your mistake, consider another approach:
Does the order matter? What if you draw two marbles and only later decide which is first and which is second? Does that alter the probability of two whites?
 
Should this not be 1/2 each?
 
The clue in this question is that you know that the second marbel is white. Imagine that when you pick the first marble, which you don't know and place in a black box, then second marble you pick is white. So now you are left with a marble in the urn and a marble in the black box. What is the probability that the marbel in the black box is white/black?

Hope this helps.
 
SrishtiSaha28 said:
Should this not be 1/2 each?
Yes.
 
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