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One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 08:30
1
Bunuel wrote:
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.
Kudos for a correct solution.
(1) The average (arithmetic mean) of the set is zero. as average is 0 and there are 11 elements , 6th element is 0 . A+(6-1)*2 =0 ; A=-10 , set is : -10,-8,-6,......0.........6,8,10 sufficient.
(2) The probability that k = 10 is the same as the probability that k = -10. P can be zero OR P can be some fraction P cannot be zero . so -10 and 10 both exists in the set and there are 11 elements. set is : -10,-8,-6,......0.........6,8,10 Sufficient.
Re: One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 10:33
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Bunuel wrote:
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.
Kudos for a correct solution.
1) The average of the set is zero. This means that our consecutive integers will be centered around zero. Listing out these numbers, we see that with 11 numbers, the only possible set is {-10,-8,-6,-4,-2,0,2,4,6,8,10}
Therefore, the probability that k = 10 = 1/11. Sufficient.
2) P(10) = P(-10).
This answer is tricky. If 10 and -10 are both in the set, then the probability of picking 10 = 1/11. However, if neither 10 nor -10 are in the set, the probability is zero. This answer is NOT sufficient.
Re: One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 11:00
Lucky2783 wrote:
Bunuel wrote:
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.
Kudos for a correct solution.
(1) The average (arithmetic mean) of the set is zero. as average is 0 and there are 11 elements , 6th element is 0 . A+(6-1)*2 =0 ; A=-10 , set is : -10,-8,-6,......0.........6,8,10 sufficient.
(2) The probability that k = 10 is the same as the probability that k = -10. P can be zero when set is set is : 12,14,16.......11elements OR P can be some fraction when set is : -10,-8,-6,......0.........6,8,10
In sufficient
answer D
overlooked the question stem and thought that 0 mean is part of question stem. Option B is insufficient.
Re: One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 11:47
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.
Re: One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 18:44
1
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
Re: One number, k , is selected at random from a set of 11 consecutive eve
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30 Mar 2015, 21:27
1
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....
Re: One number, k , is selected at random from a set of 11 consecutive eve
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31 Mar 2015, 20:37
EMPOWERgmatRichC wrote:
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
Rich, Can you give a brief over if probabilities are equal, then how could both 1/11 or 0/11 ?
Re: One number, k , is selected at random from a set of 11 consecutive eve
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01 Apr 2015, 15:40
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
Re: One number, k , is selected at random from a set of 11 consecutive eve
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