### Brain Teasers

# Choosing Marbles

Fun: (2.35)
Difficulty: (2.47)
Puzzle ID: #8808

Submitted By: lesternoronha1 Corrected By: MarcM1098

Submitted By: lesternoronha1 Corrected By: MarcM1098

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble. After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random. What is the probability that the second marble drawn is black?

### Answer

The probability is 5/8.The probability of event A happening given that event B already happened is the probability of A and B happening divided by the probability that B happened. This can be expressed as Pr(A|B)=Pr(A and B)/Pr(B).

In this case A is drawing a black marble and B is having already drawn a black marble.

Pr(A and B) = (1/2) * [(3/4)2 + (1/4)2] = 5/16.

Pr(B) = 1/2.

Pr(A|B) = Pr(A and B)/Pr(B) = (5/16)/(1/2) = 10/16 = 5/8.

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## Comments

This problem is incorrect. By placing the marble back into the bag, this makes the probabilty exactly the same as it was in the begining, and the first marble drawn has no effect on the second.

that's exactly what i said....well, thought. I don't really talk out loud to myself....

Also......you're answer is way to complicated...you shoud make it simpler for those of us that are *slow*

I agree with JenKat. This is just like flipping a coin. It does not matter if the last 100 times you got 'heads', the next flip still has a 50% chance of being 'tails' (assuming the coin is normal and not double headed or something).

The wording of the question would have to be something which asks you to take into account that the second marble is black. eg both marbles are black. The answer is 50% to the problem as stated.

The answer is accurate although terribly explained. What happens is: Probability=(3/4 X 3/4) + (1/4 X 1/4) = 10/16 = 5/8

WHY? if you choose the bag with 3 Black (B) and 1 White (W), the probablity of getting a black is 3/4...you pick one, put it back and try again, therefore your second turn is also 3/4, multiply (3/4 X 3/4)...now, add the probability of the bag containing 3 W and 1 B, which is (1/4 X 1/4).

WHY? if you choose the bag with 3 Black (B) and 1 White (W), the probablity of getting a black is 3/4...you pick one, put it back and try again, therefore your second turn is also 3/4, multiply (3/4 X 3/4)...now, add the probability of the bag containing 3 W and 1 B, which is (1/4 X 1/4).

Thanks Smooth. Makes sense to now. Looks like the answer is correct and this is a very clever stats teaser.

What he said, but good work anyways.

There is a 1 in 2 chance of selecting the bag with 3 black marbles. With that bag, there is a 3 in 4 chance of picking a black marble. 1/2 * 3/4 = 3/8. There is also a 1 in 2 chance of selecting the bag with only 1 black marble. With that bag, there is a 1 in 4 chance of picking a black marble. 1/2 * 1/4 = 1/8. 3/8 + 1/8 = 4/8 = 1/2 = 50%. After all, 4 of the 8 marbles are black.

cnmne, your logic would work if we hadn't already seen a black marble. Imagine the equivalent question but where one bag had only white in and one had only black in. After you've seen a black marble, you wouldn't still say there's a 1/2 chance of having each bag! The probabilty changes from 50-50 to absolutely certain because of knowledge you've gained.

Here, once you've seen a black marble, there's a 1/4 chance you have the bag with one black marble in and a 3/4 chance it's the other one.

Here, once you've seen a black marble, there's a 1/4 chance you have the bag with one black marble in and a 3/4 chance it's the other one.

i agree with cnmne coz i get the same answer. although you have known that you get black marble, it makes no difference coz you put it back. each bag has at least 1 black marble, so you don't know which is either the bag contains 3 black 1 white or the bag contains 1 black 3 white. the probability is 1/2*1/4+1/2*3/4=1/8+3/8=4/8=50%

No, smooth is definitely right. There are many of these puzzles where people don't take into account that we actually have more information than we think. Cnmne's logic is the correct logic for the FIRST MARBLE DRAWN. Now that we know it's black, we need to use smooth's logic. All you need to do is try it out at home about 20 times. You should pull out the SAME COLOR marble 5/8 of the time (black/black or white/white) and DIFFERENT COLOR marbles 3/8 of the time.

Smooth is right, but he forgot to divide by 2. Only HALF of the times you will choose the bag with 3 black marbles and 1 white marble(3B1W) with probability of (3/4 X 3/4) of getting 2 black marbles. The other half of the times you will choose the 1B3W bag, with probability of (1/4 X 1/4) of getting 2 blacks. The final probability is:

1/2 X [(3/4 X 3/4)+(1/4 X 1/4)]=

5/16

1/2 X [(3/4 X 3/4)+(1/4 X 1/4)]=

5/16

Question for the floor...

If there was a black marble only bag, and a white marble only bag (and you knew the distribution) and you pulled a black marble , put it backand was going to draw again, your knowledge would give you a 100% chance of pulling black, right? Then your friend came along and pulled from the same bag(he has no knowledge of what u pulled.) Does he have a 50% chance of pulling black or a 100% chance?

If there was a black marble only bag, and a white marble only bag (and you knew the distribution) and you pulled a black marble , put it backand was going to draw again, your knowledge would give you a 100% chance of pulling black, right? Then your friend came along and pulled from the same bag(he has no knowledge of what u pulled.) Does he have a 50% chance of pulling black or a 100% chance?

OK, after you take on black out here are possibilities of what is left:

3 white

2 black and 1 white.

You have 1/2 a chance of getting the bag with black left. Then, in that you have a 2/3 chance of getting a black. Therefore, it is 2/6.

3 white

2 black and 1 white.

You have 1/2 a chance of getting the bag with black left. Then, in that you have a 2/3 chance of getting a black. Therefore, it is 2/6.

Wait, I didn't consider that you put it back in...

Anyway, smooth you logic fails. In yours, your considering that it's the chance of having black and then black (which is 5/16).

But read the question again. It says:

"What are the chances of getting a black marble NEXT?"

Therefore you only have to say what chances you have next time!

Therefore, it's 50%.

But read the question again. It says:

"What are the chances of getting a black marble NEXT?"

Therefore you only have to say what chances you have next time!

Therefore, it's 50%.

smooth is correct and qwertypiusa is not.

Having drawn a black marble, the probability that you've picked the bag with three black marbles is 3/4 and that you've picked the bag with one black marble is 1/4. This information changes the probability of each bag having been selected. This is the part that many of the early comments here missed and the confusing description in the answer didn't explain. At this point replacing the marble in the bag or not doesn't change the probabilities for having selected each bag.

This gives the .75^2 + .25^2 = 5/8.

Having drawn a black marble, the probability that you've picked the bag with three black marbles is 3/4 and that you've picked the bag with one black marble is 1/4. This information changes the probability of each bag having been selected. This is the part that many of the early comments here missed and the confusing description in the answer didn't explain. At this point replacing the marble in the bag or not doesn't change the probabilities for having selected each bag.

This gives the .75^2 + .25^2 = 5/8.

Here's the answer to Spock's question above:

Information affects probability. When your friend comes up, there is a 50% chance you selected the bag with the white marbles, and 50% you selected the bag with the black marbles. YOU know his probability is 100%, but from his perspective, it is 50%.

Think of it this way. If 1000 times you select a bag at random and draw a marble out of the bag without telling your friend what it was, your friend will expect to draw a black marble around 500 times. He can't predict with any greater certainty what color marble he will pick each time even though you can predict what color he will draw each time with 100% accuracy.

Without the information you have, his probability is 50%. With your information the probability is 100%.

Information affects probability. When your friend comes up, there is a 50% chance you selected the bag with the white marbles, and 50% you selected the bag with the black marbles. YOU know his probability is 100%, but from his perspective, it is 50%.

Think of it this way. If 1000 times you select a bag at random and draw a marble out of the bag without telling your friend what it was, your friend will expect to draw a black marble around 500 times. He can't predict with any greater certainty what color marble he will pick each time even though you can predict what color he will draw each time with 100% accuracy.

Without the information you have, his probability is 50%. With your information the probability is 100%.

This is definitely a question of what is being asked... I precieved the question was the probability of picking a black marble, not the probability of picking it twice. Thus a 1/2(3/4) + 1/2(1/4)=1/2 probabillity. But if the question is picking it twice then the probability does not equal (3/4 X 3/4) + (1/4 X 1/4) = 10/16 = 5/8 as has been claimed. WHY? There is a 1/2 chance of picking each bag and in the above calculation it must be taken into account. So there is a 1/2(3/4

x 3/4) + 1/2(1/4 x 1/4) = 5/16 probability of choosing a black marble twice.

x 3/4) + 1/2(1/4 x 1/4) = 5/16 probability of choosing a black marble twice.

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