Magician has a rabbit in the hat, it's either white or black (probabil

Question

Try this one:

Magician has a rabbit in the hat, it's either white or black (probability 50/50). He adds one more rabbit in the hat which is white and then he randomly picks one which turns out to be white. What is the probability that the remaining rabbit in the hat is also white?

A. 0
B. 1/3
C. 1/2
D. 2/3
E. 1.

A. 0
B. 1/3
C. 1/2
D. 2/3
E. 1.

GMAT TIGER · Answer

The second one could be either white or black.

P (white) = white (p) + (1- black)(p) 
P (white) = 1 (1/2) + 0 (1/2) 
P (white) = 1/2

rvthryet · Answer

I think 1/2 should be the ans :oops:

AKProdigy87 · Answer

Good question. A bit tricky to get your head around it because you have to approach it using conditional probabilities in reverse to some extent. I believe the answer is  2/3 .

To Start:

Rabbit 1: 50/50 White/Black
Rabbit 2: 100 White

Now we know that the first rabbit picked was white. Therefore, there's a 2/3 chance that Rabbit 2 was picked, and 1/3 chance Rabbit 1 was picked, based on conditional probability (we know a black rabbit WAS NOT picked, so we can say 2 out of 3 times, Rabbit 2 was picked).

Therefore,

If rabbit 1 was picked initially (P = 1/3): 100% chance a white rabbit will be picked next.
If rabbit 2 was picked initially (P = 2/3): 50% chance a white rabbit will be picked next.

P(white rabbit will be picked) = (1/3)(1) + (2/3)(1/2) = 2/3

If anyone needs clarification on that, I'll try and explain in more detail. Good question, +1.

jax91 · Answer

I have a small doubt...

can we change the problem slightly to make it more simple, without changing the meaning though.

i.e. we say that the hat already had 1 black and 1 white rabbit.

so P(Black initial) = 1/2 and P(White initial) = 1/2 remain the same.

This way it becomes simpler. But has the meaning changed? :roll:

rvthryet · Answer

Let me put this question in a different way.. Imagine you have two coins

1st is a fair coin with a Head and a Tail.
2nd is a coin with only a head ( so u'll only get a head in each flip)

You have four possible outcomes in this case H, T, H, H. Out of this favourable outcomes are 3(H,H,H). So,

P(getting head in the 1st trial) = 3/4.

After the first trial only one coin is left. Since we do not know out of the two which coin is removed the possible outcomes are H,T and favourable is only 1. So,

P(getting a head in second trial)= 1/2.

Therefore, the ans to the question should be 1/2

This just seems very logical.. though i do not know if it is correct

Probability has never been my forte.. Pls let me know where am i wrong

ggarr · Answer

Do we have an OA for this?

Bunuel · Answer

Sure, but post it later. This problem is quite tricky. Try it.

ggarr · Answer

I did. I'm getting 1/2---which, at this point, I'm sure is wrong.

hgp2k · Answer

Here is what I think:

The probability that the first rabit is white = 1/2.
So after adding the second rabit, we have combination: {W, W, B}
First rabit is W, so remaining combination is: {W, B}
So again the probability that the rabit is White = 1/2.
:roll:

ggarr · Answer

I thought about this again and Prodigy's expl. makes sense to me. Things can shake out one of two ways:

Instance 1  
 w /b (scenario 1)
w/w (scenario 2)

or

Instance 2  
w/b (scenario 1)
 w /w (scenario 2)

Let's assume that the bolded Ws are the white rabbits "randomly pick " in each instance. Once the bolded W is chosen, there are 2 out of 3 chances of another W being chosen. I hope that this makes sense---and more importantly, is correct.

Bunuel · Answer

SOLUTION:

This is longknow probability problem.

After picking white rabbit there are three equally likely states:

In the hat could be:

1. White rabbit, which was there initially (picked the white we put);
2. Black rabbit, which was there initially (picked the white we put);
3. White rabbit, which we put (picked the white was there initially)

Thus probability of picking the white is 2/3.

P.S. Posted the solutions to the tough & tricky problems set at: tough-tricky-set-of-problms-85211.html#p638370

gmatjon · Answer

But, I can not follow the logic.
If problem sounds like this :

The magician has two rabbits in the hat: black and white (so their probability of being chosen is 1/2)
Then one more white rabbit is put over there, thus the probability of choosing a rabbit increased to 2/3 (probability of choosing black is 1/3)
Then white rabbit is removed. Thus only one black rabbit and one white rabbit is left, thus it means the probability should be
1/2

Is n't the problem the same, with the one you put?

AKProdigy87 · Answer

You would need to put two more white rabbits to compensate for the fact that you are treating the original rabbit as two rabbits.

amit2k9 · Answer

interesting solution.

TrueLie · Answer

I think that there is a problem with this solution because the 3 cases mentions above do not have the same probability of being true. It is not like the case we have 3 balls in a basket (2 white and 1 black). In this case the probability of picking a white ball is 2/3.

Do you have any official source of this problem?

Thank you.

SVaidyaraman · Answer

1. There could be either white or black rabbit initially. 
2. When the white rabbit is put in and later some rabbit is taken out, we have the following possibilities:

white (put in later and taken out) , white (initially present and in the hat)
white (put in later and taken out) , black (initially present and in the hat)
white(initially present and taken out), white(put in later and in the hat)
black(initially present and taken out), white(put in later and in the hat)

3.We know only a white rabbit was taken out and so disregard the last possibility.
4.Out of the first 3 possibilities , white is in the hat 2 times and black once. So the answer is 2/3.

kingbucky · Answer

There are 2 equally likely events in the sample space {WB, WW}
WB := represents the case where one black and one white rabbit are there in the hat.
WW:= represents the case where one white and one white rabbit are there in the hat.

Now,

| W B | |W| |B| 
-------- -> --------- 
| WW | |W| |W|

There is equally likely chance that we have WB and WW. Of that, we have equally likely chance that of the 1st case, the chosen rabbit is W or B. In case 2, we have equally likely W or W.
Hence given one |W| was the outcome, |WW| occurs in 2/3 of the time.

Suraj1101 · Answer

Tricky question indeed, stalled for almost an hour, still got the concept and underlying logic at the end

Let's denote the events as follows: - ( W ) is the event that the rabbit initially in the hat is white. - ( B ) is the event that the rabbit initially in the hat is black. - ( P ) is the event that a white rabbit is picked. Given that initially, the probabilities are: - P(W) = 0.5 - P(B) = 0.5 The magician adds one more white rabbit to the hat, making the scenarios: 1. If the initial rabbit was white, there are now two white rabbits in the hat. 2. If the initial rabbit was black, there is one white and one black rabbit in the hat. We are given that a white rabbit is picked. We need to find the probability that the remaining rabbit is white. Using Bayes' theorem, we need to calculate P(W | P) , the probability that the initial rabbit was white given that a white rabbit was picked. First, we calculate the probabilities of picking a white rabbit under each initial condition: - If the initial rabbit was white ( W ), the probability of picking a white rabbit is 1 because both rabbits are white. - If the initial rabbit was black ( B ), the probability of picking a white rabbit is 0.5 because there is one white and one black rabbit in the hat. Now we use Bayes' theorem: P(W | P) = (P(P | W) * P(W))/ P(P) Where: - P(P | W) is the probability of picking a white rabbit given that the initial rabbit was white. - P(W) is the prior probability of the initial rabbit being white. - P(P) is the total probability of picking a white rabbit. We already know: - P(P | W) = 1 - P(W) = 0.5 To find P(P) , use the law of total probability: P(P) = P(P | W) * P(W) + P(P | B) * P(B) Substituting the known values: P(P) = (1 * 0.5) + (0.5 * 0.5) = 0.5 + 0.25 = 0.75 Now, P(W | P) = (1 * 0.5) / 0.75 = 0.5 / 0.75 = 2/3 So, the probability that the remaining rabbit in the hat is also white, given that a white rabbit was picked, is 2/3.

Magician has a rabbit in the hat, it's either white or black (probabil

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Magician has a rabbit in the hat, it's either white or black (probabil

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