usre123 wrote:
I am not able to solve this questions with below method:
probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5
probability required = 1-2/5 = 3/5
Where am I doing wrong?
Regards,
Ravi
You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.[/quote]
Hmmm...silly again.
Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/152/5 +4/15+4/15= 14/15
so actual probabilty required is 1-14/15 = 1/15
Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?[/quote]
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A> Only Joshua(Jh) and !Jose(Jo)
so we have Joshua + Any of the (A,B,C,D)= (Jh,A) or (Jh,B) or (Jh,C) or (Jh,D) ------>[or simply we can write it as 4c1 since we have Jh already and we have to choose one from other 4]
4c1/6c2
B> Only Jo and !Jh
same 4c1/6c2
final answer =
1- (probabilty of not choosing both of them)- probability of choosing Jh but !Jo - Probability of choosing Jo but !Jh.
This is bulky and I was just checking if things would work this way...simple way is
there can only be one set with both Jo and Jh and total number of ways choosing 2 from 6 is 6c2 = 15
therefore probability is 1/15