One number, k , is selected at random from a set

[guerrero]

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.

I do not have OA right now , I will post it in few hours .

[Uva@90]

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.

I do not have OA right now , I will post it in few hours .

Hi Guerrero,
Given: We have 11 Consecutive Integers.
To Find: P(10) ?

Statement 1: The average (arithmetic mean) of the set is zero.
So 11 Numbers are,
-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10 => Avg = 0
So P(10) = 1/11
Hence Sufficient.

Statement 2: The probability that k = 10 is the same as the probability that k = -10.
We know that P(10) = P(-10)
If we have -10 in the set then P(10) will be 1/11
If we don't have -10 in the set then P(10) = 0
Hence Insufficient.

Answer is A

Regards,
Uva.

[theCodeToGMAT]

11 consecutive even integers = (A + 0) + (A+2) + (A+4) ... + (A+20) = 11A + (10)(11) = 11A + 110

To find: Probability that k = 10

Statement 1:
Average = 0
(11A + 110)/11 = 0
A = -10
We can find the probability
SUFFICIENT

Statement 2:
P(k=10) = P(k=-10)
we have no idea whether 10 & -10 are in sets.
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]

[[email protected]]

Hi guerrero,

Both Uva@90 and Rahul have correctly answered the question, but I wanted to add some details to their explanation.

Fact 1: The average of the 11 consecutive even integers is 0.

This is fairly easy to deal with; the numbers are:

-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10

The probability that K = 10 is 1/11
Fact 1 is SUFFICIENT

Fact 2: The probability that K=10 is the same as the probability that K = -10

We COULD have the same 11 numbers as we had in Fact 1 (both 10 and -10 show up), in which case the probability is 1/11

We COULD have 11 consecutive evens from 12 to 32, inclusive (neither 10 nor -10 show up), in which case the probability is 0/11
Fact 2 is INSUFFICIENT

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

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.