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Probability (m02q06)

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Probability (m02q06) [#permalink] New post 25 Oct 2008, 20:54
In preparation for the Olympics, a group of five women is participating in a 400 meter race. What is the probability that either Jenny or Sally will win this race?

(A) \frac{1}{5}
(B) \frac{4}{15}
(C) \frac{2}{5}
(D) \frac{1}{2}
(E) \frac{3}{5}

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Please provide an explanation to your answer.

When do we use the logic - A or B = P(A) + P(A) - P(A&B)
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Re: Probability [#permalink] New post 26 Oct 2008, 00:05
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1) if the question stem supposed there is only one winner then
Probab for J is 1/5
for S is 1/5
for J or S=1/5+1/5=2/5

In order to use the formula u provided two events must overlap.
ex: There 20 students. 12 are member of Drum club. 15 are members of Soccer club. If all the students are members of either club or both.What is the probability that a rendomly chosen student is a member of either club but not both.

2) But winner(s) can be 1,2,3,4,5 at the same time, why can't they cross the finish at the sametime? If this is case then it gets way complex.
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Re: Probability [#permalink] New post 26 Oct 2008, 00:09
UMB wrote:
1) if the question stem supposed there is only one winner then
Probab for J is 1/5
for S is 1/5
for J or S=1/5+1/5=2/5

In order to use the formula u provided two events must overlap.
ex: There 20 students. 12 are member of Drum club. 15 are members of Soccer club. If all the students are members of either club or both.What is the probability that a rendomly chosen student is a member of either club but not both.

2) But winner(s) can be 1,2,3,4,5 at the same time, why can't they cross the finish at the sametime? If this is case then it gets way complex.


I guess we proceed with the assumption that only one completes the race first ....
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Re: Probability [#permalink] New post 26 Oct 2008, 05:18
Excellent explanation UMB - thanks for your time.

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Re: Probability (m02q06) [#permalink] New post 11 Feb 2010, 06:49
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Hay Why cant there be four Jennys and one Sally among the five. Or Three Sallys and Two Jennnys? Etc.

The question does not specify names of all five participants....
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Re: Probability (m02q06) [#permalink] New post 11 Feb 2010, 10:44
If the question is a DS then we can pick ans choice E as you stated but it is a problem solving type of question so we should move on... and also either Jenny or Jan clearly denotes that they are just 2 out of 5 from the group... So my choice is 2/5

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Re: Probability (m02q06) [#permalink] New post 11 Feb 2010, 11:11
Any of the five women may win - five outcomes, two of them suit the condition, so 2/5.
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Re: Probability (m02q06) [#permalink] New post 12 Feb 2010, 00:22
nvgroshar wrote:
Any of the five women may win - five outcomes, two of them suit the condition, so 2/5.


No of ways in which either one of them can win = 2c1 = 2

No of ways in which one can win (total no. of outcomes) = 5c1 = 5

So P = 2c1/5c1 = 2/5
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Re: Probability (m02q06) [#permalink] New post 12 Feb 2010, 18:55
If only one of Jenny or Sally wins: 1/5 + 1/5 = 2/5

If (maybe) both win: 1/5 + 1/5 - 1/25 = 9/25

Only 2/5 is in the options, so we can safely go bo that!
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Re: Probability (m02q06) [#permalink] New post 16 Feb 2011, 13:23
lets say _ _ _ _ _ _ five places to fill.
First place should be either jenny or sally can be filled in two ways. Second place in 4 ways as one women place is fixed.
third place in 3 ways, and fourth in 2 & fifth in one way ==> 2* 4* 3* 2* 1
In general, if it is to fill five places with five women can be done in 5*4*3*2*1 ways

Probality is (2*4*3*2*1)/(5*4*3*2*1) = 2/5 C
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Re: Probability (m02q06) [#permalink] New post 16 Feb 2011, 14:40
If it was a DS question , answer would be different. Here it is clearly C).

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Re: Probability (m02q06) [#permalink] New post 16 Feb 2011, 18:52
The equation; p(a) + p(b) - p(aandb) would be used if the question had asked for the probability of j or s but not both.
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Re: Probability (m02q06) [#permalink] New post 16 Feb 2011, 21:26
ans:c 1/5+1/5 =2/5

easy one
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Re: Probability (m02q06) [#permalink] New post 16 Feb 2011, 23:52
simple one: 1/5+1/5=2/5 is ans.
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Re: Probability (m02q06) [#permalink] New post 18 Feb 2011, 06:57
Assuming only one of them can win, the probability will be 1/5 + 1/5 = 2/5.

Wow! I guess this would be one of the simpler problems of the GMAT club tests..
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Re: Probability (m02q06) [#permalink] New post 20 Feb 2012, 17:00
I got C as well, was worried that there might be a trick. Apparently not!
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Re: Probability (m02q06) [#permalink] New post 20 Feb 2012, 18:31
When a probability questions asks "or", that's usually a signal to use A or B = P(A) + P(A) - P(A&B). When the events are mutually exclusive (only one can win a race in all but the most unlikely of outcomes) P(A&B) = 0 so you're left with A or B = P(A) + P(A).
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Re: Probability (m02q06) [#permalink] New post 20 Feb 2012, 22:48
Another way to solve this question.
Since either S or J has to win the race so make them as one group so now we have 4c1
and the total no of events is 5c2
so probability is 4c1/5c2 = 2/5
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Re: Probability (m02q06) [#permalink] New post 21 Feb 2012, 04:07
Since (A&B) = 0 in this case, hence ans is P(A) + P(B)
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Re: Probability (m02q06) [#permalink] New post 20 Feb 2013, 13:09
(1/5)+(1/5)=2/5
If it was AND situation it would be multiplication.

c

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Re: Probability (m02q06)   [#permalink] 20 Feb 2013, 13:09
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