There are 20 people in a group, who are either first year student

Question

There are 20 people in a group, who are either first year student or second year student. Select 2 people randomly, is the possibility that two selected people are all first year student greater than 1/2?

(1) First year students are more than 10 people.
(2) The possibility that two selected people are all second year student is less than 1/2

amanvermagmat · Accepted Answer

Let the number of first year students = x, thus number of second year students = 20-x. Probability that both are first year students = x/20 * (x-1)/19 = (x)*(x-1) / (20*19). We have to determine if this is > 1/2. (1) x > 10. If x =11, then required probability = 11/20 * 10/19, which is < 1/2 If x =15, then required probability = 15/20 * 14/19, which is > 1/2 Not sufficient. (2) This means (20-x)/20 * (19-x)/19 < 1/2 or (20-x)*(19-x) < 190. 20-x and 19-x are two consecutive integers, whose product is < 190. If we put x=15, then 20-x and 19-x are 5 and 4 respectively, their product is < 190. And in this case the probability asked in question = 15/20 * 14/19, which is > 1/2. And if we put x=11, then 20-x and 19-x are 9 and 8 respectively, their product is also < 190. And in this case the probability asked in question = 11/20 * 10/19, which is < 1/2. Not sufficient. Combining the two statements, we can still take the same examples (of x=11 and x=15), which satisfy both the statements; but give different answers to the question asked. In case of x=11, answer is NO (probability is < 1/2) and in case of x=15, answer is YES (probability is > 1/2). So not sufficient. Hence E answer

push12345 · Answer

I) more than 10
11C2/20C2 =9/38 <1/2
15C2/20C2 =21/38 >1/2
Insufficient

II)we know probability of 2nd year 
nC2/20C2 <1/2
 n(n-1)<190

n=12 then 20-n=8 so 8*7/20*19<1/2
n=10 then 20-n=10 so probability is 1/2
Insufficient

Combining both we have multiple cases

So answer is E

Posted from my mobile device

Crytiocanalyst · Answer

The key to solving the problem is getting all the possibilities correct

1:First year students are more than 10 people.
let us assume first year students be in 11 number then the probability becomes 11/20*10/19 <0.5.....1
let us assume the first years to be 19 in number then the probability becomes 19/20*18/19 >0.5 ....2
Clearly insuff

2:The possibility that two selected people are all second year student is less than 1/2
The second and first possibility can be still considered yielding the same results

Even when 1 and 2 is combined both the possibilities exist

Hence IMO E

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There are 20 people in a group, who are either first year student

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