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A certain junior class has 1000 students and a certain

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A certain junior class has 1000 students and a certain [#permalink] New post 10 Sep 2004, 19:15
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 two students selected will be a sibling pair?

(A) 3/40000
(B) 1/3600
(C) 9/2000
(D) 1/60
(E) 1/15

This is OG quest 432. I am not 100% satisfied with the OA. I will post my logic below.

Junior Class - 1000 students - has 60 students with siblings and 940 stuents without siblings

Senior Class - 800 students - has 60 students with siblings and 740 students without siblings

Case 1 - First student selected from Junior Class

P(selecting student with sibling from J) = 60/1000

Given that a student with a sibling has been selected from Junior class,
P(selecting student with sibling from S) = 1/60

So P(selecting 2 students who are siblings) = 1/1000

Case 2 - First student selected from Senior Class

P(selecting student wtih sibling from S) = 60/800

Given that student with sibling selected from S,
P(selecting student wtih sibling from J) = 1/60

So P(selecting 2 students who are siblings) = 1/800

P(selecting 2 students who are siblings) = 2*[1/1000 + 1/800] = 9/2000
(since order of whether we select from J first or S first, does not matter)

Whats wrong with this logic? Will post the OA soon
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 [#permalink] New post 10 Sep 2004, 20:38
Hi:

When you are selecting the second student (in both the cases), u're interested in picking the first student's sibling from the entire class (Q says "Prob. of selecting a sibling pair"). There is only one such student among the entire class. Total # of favorable outcomes for the 2nd selection = # of students in entire class, since you will choose from the entire class & and not out of 60. Also, the sibling pairs are mixed in the class, u don't know which 60 belong to a sibling pair. Hence the prob of doing that will 1/(# of students in class).
Also, in this case, the answer will be independent of the sequence of selection. (Junior-Senior, Senior-Junior)

Hope this helps.

pjk
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 [#permalink] New post 11 Sep 2004, 03:13
I choose A

Prob to pick up the siblings among all the classmates J+S = 9/2000
Then probability to find the good pair = 1/60

What's the OA ?
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 [#permalink] New post 11 Sep 2004, 06:42
My reply: A
Total number of ways of selecting 1 guy from each class = 1000 x 800
Number of favourable events = 60 x 1
Pb = 60/800000 = 3/40000
(Does not matter if we choose the junior first or senior first. 60 represents the number of ways you can be lucky to get a brother in one class. 1 represents the number of you can find his brother in the other class).



Stuti, if we consider your reply.

Case 1 - First student selected from Junior Class

P(selecting student with sibling from J) = 60/1000

Given that a student with a sibling has been selected from Junior class,
<<P(selecting student with sibling from S) = 1/60 >>
P(selecting student with sibling from S) = 1/800

So P(selecting 2 students who are siblings) = 60/800000 = 3/40000


Case 2 - First student selected from Senior Class

P(selecting student wtih sibling from S) = 60/800

Given that student with sibling selected from S,
P(selecting student wtih sibling from J) = 1/1000
So P(selecting 2 students who are siblings) = 60/800000 = 3/40000


<<P(selecting 2 students who are siblings) = 2*[1/1000 + 1/800] = 9/2000
(since order of whether we select from J first or S first, does not matter)
>>

You need not add this. You add independednt events only if they are various ways of satisfying the same condition (say pb of atleast 3 girls = pb of 3 girls + pb of 4 girls + pb of five girls). Since in this case, it ends up selecting a sibling pair, you need not add these.
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 [#permalink] New post 11 Sep 2004, 11:30
hardworker_indian wrote:
My reply: A
Total number of ways of selecting 1 guy from each class = 1000 x 800
Number of favourable events = 60 x 1
Pb = 60/800000 = 3/40000
(Does not matter if we choose the junior first or senior first. 60 represents the number of ways you can be lucky to get a brother in one class. 1 represents the number of you can find his brother in the other class).



Stuti, if we consider your reply.

Case 1 - First student selected from Junior Class

P(selecting student with sibling from J) = 60/1000

Given that a student with a sibling has been selected from Junior class,
<<P(selecting student with sibling from S) = 1/60 >>
P(selecting student with sibling from S) = 1/800

So P(selecting 2 students who are siblings) = 60/800000 = 3/40000


Case 2 - First student selected from Senior Class

P(selecting student wtih sibling from S) = 60/800

Given that student with sibling selected from S,
P(selecting student wtih sibling from J) = 1/1000
So P(selecting 2 students who are siblings) = 60/800000 = 3/40000


<<P(selecting 2 students who are siblings) = 2*[1/1000 + 1/800] = 9/2000
(since order of whether we select from J first or S first, does not matter)
>>

You need not add this. You add independednt events only if they are various ways of satisfying the same condition (say pb of atleast 3 girls = pb of 3 girls + pb of 4 girls + pb of five girls). Since in this case, it ends up selecting a sibling pair, you need not add these.


Thanks hardworker...Got it. btw, OA is A
  [#permalink] 11 Sep 2004, 11:30
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