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probability
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Md.Nazrul Islam
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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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Bill@VeritasPrep
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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
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
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Bill@VeritasPrep 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.ka_t_rin wrote:Bill@VeritasPrep 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?
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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GMATGuruNY
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The probability of selecting X on the NTH PICK is equal to the probability of selecting X on the FIRST PICK.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 .
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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Bill@VeritasPrep
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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)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.
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