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by aimhigh715 » Wed Nov 28, 2007 11:55 pm
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students will be a sibling pair?

a) 3/40000
b) 1/3600
c) 9/2000
d) 1/60
e) 1/15

Read the OE, didn't really get it? My answer was c) based on 60/1000 x 60/800... why not?
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by aimhigh715 » Wed Nov 28, 2007 11:57 pm
OE was A)

Nevermind I got it... reason why it was 60/1000 x 1/800 was because of the wording.

The pair HAS to be the actual sibling pair, NOT just any 2 people in a sibling pair relationship. Thus it is 1/800 for the second ratio... !!

nevermind :)
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by camitava » Thu Nov 29, 2007 1:07 am
aimhigh715, dnt mind! I am not getting ur solution. According to the Qs, Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.
So there will be 30 students of senior class and other 30 of junior class. If so, then how come you are getting 60/1000? Again why 1/800? I am getting confused here. I think we need some more input here.
Correct me If I am wrong


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Amitava
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by sankruth » Thu Nov 29, 2007 3:35 am
camitava wrote:aimhigh715, dnt mind! I am not getting ur solution. According to the Qs, Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.
So there will be 30 students of senior class and other 30 of junior class. If so, then how come you are getting 60/1000? Again why 1/800? I am getting confused here. I think we need some more input here.
Amitava, 60 pairs mean 60 sets of siblings (so 120 in total, 60 in junior class, 60 in senior class.)

Now P(Getting a student from Junior who has a sibling in senior) = P(A) = 60/1000

Now since you have selected one student in the junior class, there will only be one student in senior class that will be his/her sibling...
P(B) = 1/800

P(A n B) = (60/1000).(1/800) = 60/800,000 = 3/400000
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by camitava » Thu Nov 29, 2007 3:38 am
Ok, sankruth! Thanks a lot. Now I understood. Sorry for my misunderstanding.
Correct me If I am wrong


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Amitava
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