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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