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# A startting line up of a team consists of x men and y women.

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Manager
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A startting line up of a team consists of x men and y women. [#permalink]

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08 Apr 2008, 11:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A startting line up of a team consists of x men and y women. There are also 4 reserve players, 2 of whom are men. If one of the starting players is unable to play and needs to be replaced by one of the reserves, what is the probability that the number of women on the starting team will increase?
1) x+y =12
2) x/y=1/3

walker can u describe it briefly plssssssssss?
.
thanks
shobuj

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08 Apr 2008, 11:10
Actually, I've done this here: 7-t58619

B

we have 4 combination:

man out, man in - the number of women is constant.
woman out, man in - the number of women decrease.
man out, woman in - the number of women increase.
woman out, woman in - the number of women is constant.

the probability that man out: $$p_{mo}=\frac{x}{x+y}$$

the probability that woman in: $$p_{wi}=\frac{2}{4}$$

the probability that the number of women increase: $$p=p_{mo}*p_{wi}=\frac{x}{x+y}*\frac{2}{4}=\frac{1}{1+\frac{x}{y}}*\frac{1}{2}$$

only second condition gives us $$\frac{x}{y}$$
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Manager
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08 Apr 2008, 11:51
THANKS
WALKER
VERY VERY MUCH

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Manager
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08 Apr 2008, 12:14
WALKER I REALLY APPRECIATE UR DESCRIPTION
BUT PLS CAN U DEFINE THIS

I DONT UNDERSTAND THE 3RD LINE OF UR EQUATION X/(x+y)*(2/4)=(1/1+(X/Y))*1/2

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08 Apr 2008, 13:06
I just wrote the probability as a product of two probabilities. I also showed that we do not have to know x and y, just x/y.

If it is not clear, tell me what exactly you don't understand.
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11 Apr 2008, 11:25
how u go from here X/(x+y)*(2/4) to there 1/1+(X/Y))*1/2
can u show it clearly

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11 Apr 2008, 11:42
shobuj wrote:
how u go from here X/(x+y)*(2/4) to there 1/1+(X/Y))*1/2
can u show it clearly

$$p=\frac{x}{x+y}*\frac{2}{4}=\frac{\frac{x}{x}}{\frac{x}{x}+\frac{y}{x}}*\frac{1}{2}=\frac{1}{1+\frac{x}{y}}*\frac{1}{2}$$
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11 Apr 2008, 22:44
We want to know what the odds are for a man out being replaced by a woman. We know the odds that a woman will be chosen if we know the female male ratio for replacements, which is 1/2 in this case. We will also know the odds that a man is sick if we know the female male ratio for startups. That's all we need to know to solve this problem. We don't even need to write the formulas out since we are not asked to give an answer. We just need to tell them if an answer could be obtained under which condition. This way you save a little time again.
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12 Apr 2008, 03:34
walker wrote:
man out, man in - the number of women is constant.
woman out, man in - the number of women decrease.
man out, woman in - the number of women increase.
woman out, woman in - the number of women is constant.

the probability that man out: $$p_{mo}=\frac{x}{x+y}$$

the probability that woman in: $$p_{wi}=\frac{2}{4}$$

the probability that the number of women increase: $$p=p_{mo}*p_{wi}=\frac{x}{x+y}*\frac{2}{4}=\frac{1}{1+\frac{x}{y}}*\frac{1}{2}$$

only second condition gives us $$\frac{x}{y}$$

walker, why should we multiply two probabilities? why not to add them? is it rule?

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12 Apr 2008, 03:52
kazakhb wrote:
walker, why should we multiply two probabilities? why not to add them? is it rule?

Yes, It is a rule for two independent probabilities.

For example, we throw a coin and a dice. What is probability to have a head for the coin and "2" for the dice?

We have two independent events and the probability to have two events is a product of two probability for each event:

p=1/2 * 1/6 = 1/12
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12 Apr 2008, 04:09
walker wrote:
kazakhb wrote:
walker, why should we multiply two probabilities? why not to add them? is it rule?

Yes, It is a rule for two independent probabilities.

For example, we throw a coin and a dice. What is probability to have a head for the coin and "2" for the dice?

We have two independent events and the probability to have two events is a product of two probability for each event:

p=1/2 * 1/6 = 1/12

sorry for so many questions
how can we differentiate whether events are dependent or independent? Also I don't see difference between independent event and mutual exclusive events, are they same?

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12 Apr 2008, 04:49
first event: a dice gives "1"
second event: a dice gives "2"

If we are talking about the same dice, the dice cannot give 1 and 2 simultaneously (mutual exclusive, when one event happens, the other event cannot happen). So, the probability to have 1 or 2 is 1/6+1/6

If we have two different dice (independent, the result of fist event does not depend on result of second event) and the probability to have 1 for one dice and 2 for other dice is 1/6*1/6
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12 Apr 2008, 08:42
walker wrote:
shobuj wrote:
how u go from here X/(x+y)*(2/4) to there 1/1+(X/Y))*1/2
can u show it clearly

$$p=\frac{x}{x+y}*\frac{2}{4}=\frac{\frac{x}{x}}{\frac{x}{x}+\frac{y}{x}}*\frac{1}{2}=\frac{1}{1+\frac{x}{y}}*\frac{1}{2}$$

Why did you switch the x and y in the last result? Shouldn't it be y/x instead of x/y?
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12 Apr 2008, 09:25
chineseburned wrote:
Why did you switch the x and y in the last result? Shouldn't it be y/x instead of x/y?

Thanks for typo
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20 Mar 2009, 13:48
walker wrote:
Actually, I've done this here: 7-t58619

B

we have 4 combination:

man out, man in - the number of women is constant.
woman out, man in - the number of women decrease.
man out, woman in - the number of women increase.
woman out, woman in - the number of women is constant.

the probability that man out: $$p_{mo}=\frac{x}{x+y}$$

the probability that woman in: $$p_{wi}=\frac{2}{4}$$

the probability that the number of women increase: $$p=p_{mo}*p_{wi}=\frac{x}{x+y}*\frac{2}{4}=\frac{1}{1+\frac{x}{y}}*\frac{1}{2}$$

only second condition gives us $$\frac{x}{y}$$

wouldn't the other three combinations affect the probability of the event?

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Re: probability ?   [#permalink] 20 Mar 2009, 13:48
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