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03 May 2018, 21:46
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E
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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
(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
03 May 2018, 22:52
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Hungluu92vn
There are 20 people in a group, who are either first year student or second year student. Select 2 people randomly, is the possilibility 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
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
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
General Discussion
07 May 2018, 09:17
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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
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
