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probability

by Md.Nazrul Islam » Fri Mar 30, 2012 6:28 pm
A bag contain 3 white balls ,3 black balls ,and 2 red balls .one by one three balls are drawn without replacement .what is the probability that the third ball is red .

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by [email protected] » Fri Mar 30, 2012 7:27 pm
First off, since the question specifically asks about drawing a red ball, we can ignore the fact that we're given 3 colors (black, white, red) and look at the problem as an either/or situation: the ball is either red or not red.

How can we get a red ball on the third draw? Three ways:

NR-NR-R: 6/8 * 5/7 * 2/6 = 10/56
NR-R-R: 6/8 * 2/7 * 1/6 = 1/28
R-NR-R: 2/8 * 6/7 * 1/6 = 1/28

1/28 + 1/28 + 10/56 = 2/56 + 2/56 + 10/56 = 14/56.

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by ka_t_rin » Thu Apr 05, 2012 2:47 am
[email protected] wrote:First off, since the question specifically asks about drawing a red ball, we can ignore the fact that we're given 3 colors (black, white, red) and look at the problem as an either/or situation: the ball is either red or not red.

How can we get a red ball on the third draw? Three ways:

NR-NR-R: 6/8 * 5/7 * 2/6 = 10/56
NR-R-R: 6/8 * 2/7 * 1/6 = 1/28
R-NR-R: 2/8 * 6/7 * 1/6 = 1/28

1/28 + 1/28 + 10/56 = 2/56 + 2/56 + 10/56 = 14/56.

Bill

Can you tell me why I got another result?
Desired outcomes:
WWR BBR WBR BWR BRR WRR RBR RWR - 8 OUTCOMES

All outcomes 3P8 = 56 thus probability is 8/56 = 1/7

What is my mistake?

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by ka_t_rin » Thu Apr 05, 2012 2:55 am
ka_t_rin wrote:
[email protected] wrote:First off, since the question specifically asks about drawing a red ball, we can ignore the fact that we're given 3 colors (black, white, red) and look at the problem as an either/or situation: the ball is either red or not red.

How can we get a red ball on the third draw? Three ways:

NR-NR-R: 6/8 * 5/7 * 2/6 = 10/56
NR-R-R: 6/8 * 2/7 * 1/6 = 1/28
R-NR-R: 2/8 * 6/7 * 1/6 = 1/28

1/28 + 1/28 + 10/56 = 2/56 + 2/56 + 10/56 = 14/56.

Bill

Can you tell me why I got another result?
Desired outcomes:
WWR BBR WBR BWR BRR WRR RBR RWR - 8 OUTCOMES

All outcomes 3P8 = 56 thus probability is 8/56 = 1/7

What is my mistake?
I believe I realized what was wrong.
To count desired outcomes we can use slot method starting from the most restricted: 2 for red, 7 and 6 for two others. = 2x7x6
and combinations for all results = 3C8 = 6x7x8
Thus 2x7x6/6x7x8 = 14/56

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by raunekk » Thu Apr 05, 2012 3:17 am
@Bill

If the question would have mentioned - Red ONLY at the third time (not the first and second time), then the answer would have been :

NR-NR-R: 6/8 * 5/7 * 2/6 = 10/56

right?

Thanks in advance.
Last edited by raunekk on Thu Apr 05, 2012 4:11 am, edited 1 time in total.

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by GMATGuruNY » Thu Apr 05, 2012 3:58 am
Md.Nazrul Islam wrote:A bag contain 3 white balls ,3 black balls ,and 2 red balls .one by one three balls are drawn without replacement .what is the probability that the third ball is red .
The probability of selecting X on the NTH PICK is equal to the probability of selecting X on the FIRST PICK.

On the FIRST pick, P(R) = 2/8 = 1/4.
Thus, on the THIRD PICK, P(R) = 1/4.
The two probabilities are equal.

If the problem asked for the probability that the FOURTH marble is red, the answer would be the same: 1/4.
If the problem asked for the probability that the FIFTH marble is red, the answer would be the same: 1/4.

For other problems that explore this concept, check here:

https://www.beatthegmat.com/manhattan-pr ... 89481.html

https://www.beatthegmat.com/probablity-ques-t60161.html
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by [email protected] » Thu Apr 05, 2012 6:53 am
raunekk wrote:@Bill

If the question would have mentioned - Red ONLY at the third time (not the first and second time), then the answer would have been :

NR-NR-R: 6/8 * 5/7 * 2/6 = 10/56

right?

Thanks in advance.
Yup. We need one of the 6 not-reds on the first draw (out of 8 possibilities, one of the 5 remaining not-reds on the second draw (out of 7 possibilities), and one of the 2 reds on the third draw (out of 6 possibilities)
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