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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?

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.

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 _________________

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.

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

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.

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[ed]" 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.

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)

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? _________________

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?

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

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)

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.

gmatclubot

Re: Magician with the rabits
[#permalink]
04 May 2011, 02:13

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