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Probability (m02q06) [#permalink]
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?

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.

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 .... _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: Probability (m02q06) [#permalink]
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

Re: Probability (m02q06) [#permalink]
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

Re: Probability (m02q06) [#permalink]
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).

Re: Probability (m02q06) [#permalink]
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