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by alex.gellatly » Tue Apr 17, 2012 4:41 am
A certain junior class has 1,000 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 selected will be a sibling pair?

(A) 3 / 40,000
(B) 1 / 3,600
(C) 9 / 2,000
(D) 1 / 60
(E) 1 / 15

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by Anurag@Gurome » Tue Apr 17, 2012 4:47 am
alex.gellatly wrote:A certain junior class has 1,000 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 selected will be a sibling pair?

(A) 3 / 40,000
(B) 1 / 3,600
(C) 9 / 2,000
(D) 1 / 60
(E) 1 / 15

Thanks
Required probability = No. of favorable outcome/ Total no. of outcomes
Total no. of outcomes = 800C1 * 1000C1
There are 60 sibling pairs, so no. of favorable outcomes = 60
Therefore, Required probability = = 60/(800*1000) = [spoiler]3/40,000[/spoiler]

The correct answer is A.
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by aneesh.kg » Tue Apr 17, 2012 4:56 am
Probability of an event = (No. of favourable outcomes for the event)/ (Total no. of possible outcomes)

The phrase 'If 1 student is to be selected at random from each class' in the question tells us about the Total no. of possible outcomes.

So, Total number of outcomes = 1000C1 * 800C1 = 1000*800

The phrase 'what is the probability that the 2 students selected will be a sibling pair?' talks about the favourable outcomes. Those outcomes in which a sibling pair is selected is favourable to us.

So, Total number of favourable outcomes = 60C1 = 60

Therefore,

Required Probability = 60 / (1000*800) = 3 / 40000

Option (A) is correct.
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by spartacus1412 » Wed Apr 18, 2012 8:11 pm
Hi,
I tied to solve the question as below:
probablity of selecting a pair = prob. of 1 out of 60 students from 1000 batch and selecting its sibling pair from the 800 student batch and vice -versa.

i.e P= (60/800)*(1/1000)+(1/800)*(60/1000)
what is wrong in this?
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by sam2304 » Wed Apr 18, 2012 9:27 pm
spartacus1412 wrote:Hi,
I tied to solve the question as below:
probablity of selecting a pair = prob. of 1 out of 60 students from 1000 batch and selecting its sibling pair from the 800 student batch and vice -versa.

i.e P= (60/800)*(1/1000)+(1/800)*(60/1000)
what is wrong in this?
You are considering selecting a senior/junior and junior/senior to be different cases, that is the problem.

Lets say we have two persons A and B, who are siblings. The question asks for a sibling pair, now selecting A and B is no different from selecting B and A. Both are same - it comprises one possibility of selecting a sibling pair.
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by GMATGuruNY » Thu Apr 19, 2012 3:08 am
alex.gellatly wrote:A certain junior class has 1,000 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 selected will be a sibling pair?

(A) 3 / 40,000
(B) 1 / 3,600
(C) 9 / 2,000
(D) 1 / 60
(E) 1 / 15

Thanks
Junior class:
Of the 1000 students, 60 are members of sibling pair.
P(selecting a member of a sibling pair) = 60/1000 = 3/50.

Senior class:
At this point, there is only ONE favorable outcome: selecting the sibling of the junior already chosen.
P(selecting the junior's sibling) = 1/800.

Since both events must happen in order to get a favorable outcome, we multiply the fractions:
3/50 * 1/800 = 3/40,000.

The correct answer is A.
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