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Given a bag with 2 black balls and 3 white balls [#permalink]

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18 Jan 2012, 02:23

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Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

Hi, struggling with this question. Your explanations would be very much appreciated!

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3 2. 1/5 3. 3/4 4. 1/2 5. 3/5

This is a conditional probability question. Basically we are told that out of 2 balls picked one is white and are asked to find the probability of the other one also to be white.

So, possible options for this other ball would be: W, W, B, B (as one white is already picked). Hence the probability for that ball to be white too is 2/4=1/2.

Hi, struggling with this question. Your explanations would be very much appreciated!

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3 2. 1/5 3. 3/4 4. 1/2 5. 3/5

Let me add a bit to what Bunuel said above:

Probability of event A = # of outcomes when A occurs/Total no of outcomes

When you have a condition given to you, the total no of outcomes adjust to that.

When you pick two balls, you can pick: {W, W}, {W, B}, {B, W} or {B, B}

But in this question, the total number of outcomes is not 4. We are given that the second ball is white. So the total number of outcomes is 2. Only {W, W} and {B, W} are possible.

What is the probability that the first one was also white? This happens in only 1 outcome - {W, W}

Hi, struggling with this question. Your explanations would be very much appreciated!

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3 2. 1/5 3. 3/4 4. 1/2 5. 3/5

Let me add a bit to what Bunuel said above:

Probability of event A = # of outcomes when A occurs/Total no of outcomes

When you have a condition given to you, the total no of outcomes adjust to that.

When you pick two balls, you can pick: {W, W}, {W, B}, {B, W} or {B, B}

But in this question, the total number of outcomes is not 4. We are given that the second ball is white. So the total number of outcomes is 2. Only {W, W} and {B, W} are possible.

What is the probability that the first one was also white? This happens in only 1 outcome - {W, W}

Probability = 1/2

Though this approach worked for this particular problem it might fail for others: suppose we have 10 white and 2 black balls and need to find the probability that the second ball is white while knowing that the first ball we picked was also white.

We also would have only two possible scenarios {W, W} and {W, B} but here we can not say that the answer is 1/2, because {W, W} and {W, B} will have the different # of ways to occur.

So, the answer for the above question would be 9/11 and not 1/2.
_________________

Re: Given a bag with 2 black balls and 3 white balls [#permalink]

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12 Jul 2013, 19:37

Let w1, w2, w3, b1 and b2 be the balls. 1. The possible outcomes : w1b1 w1 b2 w2 b1 w2b2 w3 b1 w3 b2 b1 w1 b2 w1 b1w2 b2 w2 b1w3 b2w3 w1 w2 w1 w3 w2 W3 w3 w1 w3 w2 w2 w1 b1 b2 b2 b1

But the total number of outcomes. is actually the number of times the second ball is white and the favorable outcomes is the number of times in that the first ball is also white.

2. The second ball is white on 12 occasions 3. The first ball is also white on 6 occasions 4. So the required probability is 6/12 = 1/2

Alternate approach:

1. we do not know what color is the first ball. Let it be of color x 2. The second ball is white 3. the 2 selections are x,W 4. Since we know the second ball is white, the first ball should have been filled by one among 2 white balls and 2 black balls 5. The probability of that being a white ball is 2/4=1/2
_________________

Re: Given a bag with 2 black balls and 3 white balls [#permalink]

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17 Aug 2013, 02:36

My approach is this:

Probability = No. of favorable cases/ total no. of cases

Total No. of cases = no. of times the second ball drawn is white. No. of favorable cases = No. of cases in which the first ball drawn is also white given that the second ball drawn is white

Req. Probab. = (Prob. of drawing a white ball in the first and second attempt)/{(Prob. of drawing a white ball in the first and second attempt)+(Prob. of drawing a black ball in the first attempt and a white ball in the second attempt)} Hence required probability = {(3/5)*(2/4)}/{(3/5)*(2/4) + (2/5)*(3/4)} = 1/2

Re: Given a bag with 2 black balls and 3 white balls [#permalink]

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17 Aug 2013, 03:23

Splendidgirl666 wrote:

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3 2. 1/5 3. 3/4 4. 1/2 5. 3/5

The probability of the 2nd will not affect the 1st any how. The probability should be the normal one.

or the question is ambiguous. i would like to know about the source because of the voting result i have seen here.....
_________________

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3 2. 1/5 3. 3/4 4. 1/2 5. 3/5

The probability of the 2nd will not affect the 1st any how. The probability should be the normal one.

or the question is ambiguous. i would like to know about the source because of the voting result i have seen here.....

And again, there is nothing wrong with the question.
_________________

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