(1) Either 5 or 6 or 7 members are older than 35 years. Probability is minimum if it is 5. For 5 the probability is C(5, 2)/C(8, 2) = 5/14 = 15/42 > 14/42 = 1/3. SUFFICIENT

(2) P(Both ≤ 35) > 1/10. There are other cases too. Like one of them less than 35, another greater and vice versa. No info about that. INSUFFICIENT.

Got the same answer, but arrived at it differently.

Statement 1:

First of all, if more than half of 8 members are older than 35, that means that 5,6,7 or all members of the group are older than 35. Let's take the case of the least number of old people, 5.

The probability of selecting 2 members older than 35 is then: 5/8*4/7=20/56 which is bigger than 1/3.

For all other possible cases, the result will be even bigger than 20/56, so that it is possible to say that p>1/3.

SUFFICIENT

Statement 2:

Let's say 2 members of the group are younger than 35. That means the probability of selecting two such members is 2/8*1/7=2/14 which is bigger than 1/10.

By extension, we then have 6 members older than 35. The probability of selecting two such members would then be 6/8*5/7=30/56 which is definitely bigger than 1/3.

Now let's take a different scenario. Assume we have 6 members younger than 35. According to what I said above, the probability of selecting two young members is definitely > 1/10.

In that case the probability of selecting two members older than 35 would decrease to 2/8*1/7=2/56 which is smaller than 1/3.

INSUFFICIENT.

The answer is A.





Oddball, I'm not sure about the formula you are using for statement 1. Could you please explain that? If never seen such a formula. Thanks.