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Probability DS: BTG PQ

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Probability DS: BTG PQ

by The Jock » Sun Sep 12, 2010 9:14 pm
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 am not sure how to deduce the solution from the second option. Please help
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Varun
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by manishankar_83 » Mon Sep 13, 2010 2:48 am
What's the OA, according to me the answer should be A

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by Maciek » Mon Sep 13, 2010 6:54 am
Hi all!

IMO A

There are 11 consecutive even integers in a set. The probability of selecting 10 at random is:
either equal to 0, if 10 is not in a set,
or equal to 1/11, if 10 is in a set.

(1) arithmetic mean = 0
the formula for arithmetic mean of the set:
mean = (I1 + I2 + ... I11)/11
If the set consists of odd number of consecutive even integers, average mean must be equal to middle integer.
Therefore, the set includes 10. S = {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}.
The probability is equal to 1/11.
The statement (1) ALONE is SUFFICIENT.

(2) The probability that k = 10 is the same as the probability that k = -10.
It means that
either 10 and -10 both are in the set
or 10 and -10 both are not in the set
The statement (2) ALONE is INSUFFICIENT.

Hope it helps!
Best,
Maciek
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