Hi guys,
Although I did arrive at the right answer I am not so sure of my method. Can someone tell me how to solve this. Thanks!
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 selcted 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
Probability!!
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For two siblings to be selected, 2 things must happen: we must select a junior who has a senior sibling AND the senior selected must be the sibling of the selected junior.RSK wrote:Hi guys,
Although I did arrive at the right answer I am not so sure of my method. Can someone tell me how to solve this. Thanks!
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 selcted 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
We get P(junior with sibling AND selected senior is sibling to selected junior)= P(junior with sibling) x P(selected senior is sibling to selected junior)
P(junior with sibling): there are 1000 juniors and 60 of them have senior siblings. So, P(junior with sibling)=60/1000
P(selected senior is sibling to selected junior): Once the junior has been selected, there is only 1 senior (out of 800 seniors) who is the sibling to the selected junior. So, P(selected senior is sibling to selected junior)= 1/800
So, the probability is (60/1000)x(1/800) = 3/40,000
The answer is A
Cheers,
Brent
GMAT/MBA Expert
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Perfect - great work!RSK wrote:Thank you Brent!
I did it this way: Total no. of outcomes = 1000 x 800
Favourable outcomes = 60 x 1
Probability = 60/800000 = 3/40000
Is this method correct?
That's the great thing about probability questions: You often have the option of applying counting techniques or using rules of probability.
Cheers,
Brent