One number, k , is selected at random from a set of 11 consecutive eve

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

Hi countonhearts,

You have to be careful with this question. Fact 2 states that the probabilities are equal; that means that could both be 1/11 OR they could both be 0/11....

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

MAGOOSH OFFICIAL SOLUTION:

overlooked the question stem and thought that 0 mean is part of question stem.
Option B is insufficient.

Option A is the right answer

Answer A. Statement 1 alone is sufficient.

Statement 1 and the fact that 11 consecutive numbers are present implies that the numbers are {-10,-9,...,0,1,...,10} So, we can find probability for k=10.

Statement 2. Here there are two possibilities. Either -10 and 10 are both present in the set, OR -10 and 10 are both absent from the set. In both cases probability will be different. Hence not sufficient.

Answer D.

Statement1 . Suff
AM = 0, that shows ---------------------------------
-10 -8 -6 -4 -2 0 2 4 6 8 10
Total = 11.
Probability of K to be selected. 1/11

Statement 2: P(10)= P(-10 )
11 even consecutive terms. draws the same condition as mentioned in statement 2.

Hence D

Rich,
Can you give a brief over if probabilities are equal, then how could both 1/11 or 0/11 ?

Thanks

Hi countonhearts,

Both of these examples were included in my original explanation. Note in the first example how BOTH 10 and -10 are in the set; in the second example, BOTH 10 and -10 are NOT in the set.....

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

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

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