One number, k , is selected at random from a set of 11 consecutive even integers. What is the probability that
k = 10?
(1) The average (arithmetic mean) of the set is zero.
(2) The probability that k = 10 is the same as the probability that k = -10.
If 10 is in the set, P(k is 10) = 1/11.
If 10 is not in the set, P(k is 10) = 0.
We need to know whether 10 is in the set.
Statement 1: average = 0
When numbers are EVENLY SPACED, average = median.
Thus, the median of the 11 consecutive even integers is 0, yielding the following set:
{-10, -8, -6...0...6, 8, 10}.
Since 10 is in the set, P(k is 10) = 1/11.
SUFFICIENT.
Statement 2: P(k is -10) = P(k is 10)
Case 1: The 11 consecutive even integers are {-10, -8, -6....6, 8, 10}.
Here, P(k is -10) = P(k is 10) = 1/11.
Case 2: The 11 consecutive even integers are {12, 14, 16...28, 30, 32}.
Here, P(k is -10) = P(k is 10) = 0.
INSUFFICIENT.
The correct answer is
A.
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