Hi All,
The prompt tells us that we're dealing with 11 consecutive even integers. We're then asked for the probability that a randomly selected number (K) from this set will equal 10. Based on this set-up, there are only two possibilities:
IF the number 10 is in the group, then the answer is 1/11
IF the number 10 is NOT in the group, then the answer is 0/11
So the real 'focus' of this question is about determining whether the number 10 is conclusively in the set (or not) or is possibly in the set (but possibly not).
Fact 1: The average (arithmetic mean) of the set is zero.
Since the numbers in the set are consecutive, knowing the average allows us to figure out the specific values of the numbers. Here, those numbers are...
-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10
The number 10 is conclusively here, so the answer to the question is 1/11
Fact 1 is SUFFICIENT.
Fact 2: The probability that K = 10 is the same as the probability that K = -10.
There are 2 possibilities here.
IF....
Both -10 and 10 are in the set, as they are in this example:
-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10
The answer to the question is 1/11
IF....
Both -10 and 10 are NOT in the set, as is the case in this example:
12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32
The answer to the question is 0/11
Fact 2 is INSUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
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