I disagree with the beatthegmat answer to this DS question

[peco]

Please help clarify the solution to this question. Thanks.

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.


The answer given by BTG is A

Thanks.

[shovan85]

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.

When 11 consecutive even integers AM will be 0, If the sum of all the integers equals to zero.

This means the sequence is
-10 -8 -6 -4 -2 0 2 4 6 8 10

Thus the probability of k = 10 is 1/11. Sufficient.

(2) The probability that k = 10 is the same as the probability that k = -10.

So if the sequence is -10 -8 -6 -4 -2 0 2 4 6 8 10
then P(k = 10) = 1/11 and P(k = -10) = 1/11

But say if the sequence is something else (say 12 14 16....30 32).
n this sequence P(k = 10) = 0 and P(k = -10) = 0.

Thus Not Sufficient.

IMO A

[Rahul@gurome]

Please help clarify the solution to this question. Thanks.

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.


The answer given by BTG is A

Thanks.

The answer is A indeed.
May I know what was your approach?
Here is the solution...

Given: One number, k is to be selected at random from a set of 11 consecutive even integers. Therefore the probability of k being a particular even integer is either 1/11 (in case the integer is included in the set) or 0 (in case the integer is not included).

Statement 1: The average (arithmetic mean) of the set is zero.
Only possible set of 11 consecutive even integers such that their average is zero is {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}. Thus, 10 is included in the set.

Sufficient.


Statement 2: The probability that k = 10 is the same as the probability that k = -10.
The probability may be 1/11 in case 10 and -10 both are present in the set. For example, {-10 , -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}. Or the probability may be zero in case 10 and -10 both are not present in the set. For example, {20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40}.

Not sufficient.

The corrcet answer is A.

[peco]

okay, now i think i understand the logic. i was initially of the opinion that another consecutive set of integers could give an average of zero using a set (-20,-16,-12,-8,-4,0,4,8,12,16,20) but i have just realised that this defeats the purpose of a consecutive set.

The explanation is very helpful.

Thanks a lot for clarifying. Now i feel very bad as i didnt understand the question stem before concluding.

[rishab1988]

The answer should be A.

Here is my different approach.

Let the first term in the sequence =x
We know it is sequence of even integers so the next term (2) = x+2
Similarly the last term = x+20

In an A.P average = average of first and last term = (x+x+20)/2=x+10

mean(avg)=0
x=-10

last term = x+20=10

since there are 11 terms in the sequence and only 1 of them is 10.the probability =1/11.

sufficient

2) probability of -10 is same as +10.

The first case scenario is that both have probability is =1/11 [when mean =0]

But what if both have 0 probability?

the sequence could start from 12 and go on till 32... a valid sequence of 3 numbers.

Hence,insufficient

Hence the answer should be A