*source:- "BTG practice Question"
Hi All,
Please help in the following question. However, i was not able to solve the second option.
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.
Answer is A
Looking forward to the reply
Regards
Saurabh
BTG practice Question - 700 level (probability)
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- saurabhkamal1981
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If 10 is in the set, P(k=10) = 1/11.saurabhkamal1981 wrote:*source:- "BTG practice Question"
Hi All,
Please help in the following question. However, i was not able to solve the second option.
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.
Answer is A
If 10 is not in the set, P(k=10) = 0.
We need to know whether 10 is in the set.
Statement 1: average = 0
Since average = sum/number, if the average = 0, then the sum = 0.
The 11 consecutive even integers thus must be: -10, -8, -6....6, 8, 10.
P(k=10) = 1/11.
Sufficient.
Statement 2: P(k=-10) = P(k=10)
The 11 consecutive even integers could be: -10, -8, -6....6, 8, 10.
P(k=-10) = P(k=10) = 1/11.
The 11 consecutive even integers could be: 12, 14, 16...28, 30, 32.
P(k=-10) = P(k=10) = 0
Insufficient.
The correct answer is A.
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- saurabhkamal1981
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Hi Mitch,GMATGuruNY wrote:
Statement 2: P(k=-10) = P(k=10)
The 11 consecutive even integers could be: -10, -8, -6....6, 8, 10.
P(k=-10) = P(k=10) = 1/11.
The 11 consecutive even integers could be: 12, 14, 16...28, 30, 32.
P(k=-10) = P(k=10) = 0
Insufficient.
The correct answer is A.
Thanks for your reply. Can you please elaborate the second option.
The second statement simply says " The probability that k = 10 is the same as the probability that k = -10". So, this statement satisfies the question if "The 11 consecutive even integers could be: -10, -8, -6....6, 8, 10.
P(k=-10) = P(k=10) = 1/11".
But, why this " The 11 consecutive even integers could be: 12, 14, 16...28, 30, 32.
P(k=-10) = P(k=10) = 0
Insufficient ?
In the first statement it is simply given "The average (arithmetic mean) of the set is zero". In order to get this we know "Since average = sum/number, if the average = 0, then the sum = 0.
The 11 consecutive even integers thus must be: -10, -8, -6....6, 8, 10.
P(k=10) = 1/11.
Sufficient"
Why the same cannot be true for the second statement ? It is given that "k = 10 is the same as the probability that k = -10". So, even consecutive integers should be -10, -8, -6, .....6, 8, 10.
Am i missing something ? Please help.
Waiting for your reply Mitch.
Regards
Saurabh
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If P(10) can be only one value, the statement is sufficient.saurabhkamal1981 wrote:Hi Mitch,GMATGuruNY wrote:
Statement 2: P(k=-10) = P(k=10)
The 11 consecutive even integers could be: -10, -8, -6....6, 8, 10.
P(k=-10) = P(k=10) = 1/11.
The 11 consecutive even integers could be: 12, 14, 16...28, 30, 32.
P(k=-10) = P(k=10) = 0
Insufficient.
The correct answer is A.
Thanks for your reply. Can you please elaborate the second option.
The second statement simply says " The probability that k = 10 is the same as the probability that k = -10". So, this statement satisfies the question if "The 11 consecutive even integers could be: -10, -8, -6....6, 8, 10.
P(k=-10) = P(k=10) = 1/11".
But, why this " The 11 consecutive even integers could be: 12, 14, 16...28, 30, 32.
P(k=-10) = P(k=10) = 0
Insufficient ?
In the first statement it is simply given "The average (arithmetic mean) of the set is zero". In order to get this we know "Since average = sum/number, if the average = 0, then the sum = 0.
The 11 consecutive even integers thus must be: -10, -8, -6....6, 8, 10.
P(k=10) = 1/11.
Sufficient"
Why the same cannot be true for the second statement ? It is given that "k = 10 is the same as the probability that k = -10". So, even consecutive integers should be -10, -8, -6, .....6, 8, 10.
Am i missing something ? Please help.
Waiting for your reply Mitch.
Regards
Saurabh
If P(10) can be more than one value, the statement is insufficient.
Only one set of values satisfies statement 1: {-10, -8, -6...6, 8, 10}.
Thus, P(10) can be only one value: P(10) = 1/11.
Sufficient.
But more than one set of values satisfies statement 2.
The set could be {-10, -8, -6...6, 8, 10}. In this case, P(10) = 1/11.
The set could be {12, 14, 16...28, 30, 32}. In this case, P(10) = 0.
Since P(10) can be more than one value, insufficient.
Clear?
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- saurabhkamal1981
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Thank you very much for your reply Mitch. Really appreciate for your help. To be honest i am confused but i am going to take this as learning and going to analyze this with other questions.GMATGuruNY wrote:
If P(10) can be only one value, the statement is sufficient.
If P(10) can be more than one value, the statement is insufficient.
Only one set of values satisfies statement 1: {-10, -8, -6...6, 8, 10}.
Thus, P(10) can be only one value: P(10) = 1/11.
Sufficient.
But more than one set of values satisfies statement 2.
The set could be {-10, -8, -6...6, 8, 10}. In this case, P(10) = 1/11.
The set could be {12, 14, 16...28, 30, 32}. In this case, P(10) = 0.
Since P(10) can be more than one value, insufficient.
Clear?
Regards
Saurabh
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If I may interject, what Mitch is trying to say is that the question stem allows two possible values for the probability of k=10: either 1/11 (if the set includes 10), or 0 (if the set does not include 10).saurabhkamal1981 wrote:Thank you very much for your reply Mitch. Really appreciate for your help. To be honest i am confused but i am going to take this as learning and going to analyze this with other questions.GMATGuruNY wrote:
If P(10) can be only one value, the statement is sufficient.
If P(10) can be more than one value, the statement is insufficient.
Only one set of values satisfies statement 1: {-10, -8, -6...6, 8, 10}.
Thus, P(10) can be only one value: P(10) = 1/11.
Sufficient.
But more than one set of values satisfies statement 2.
The set could be {-10, -8, -6...6, 8, 10}. In this case, P(10) = 1/11.
The set could be {12, 14, 16...28, 30, 32}. In this case, P(10) = 0.
Since P(10) can be more than one value, insufficient.
Clear?
Regards
Saurabh
So if stat. (2) says that the probability k=10 is the same as the probability of k=-10, all we can really learn from this is that 10 and -10 are both inside the set or both outside the set together: Either p(10) = p(-10) = 1/11 (both are in the set), or p(10) = p(-10) = 0 (if both are out of the set). But we still have two possible answers as to the probability of k=10, which is why stat. (2) is insufficient.
- saurabhkamal1981
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Thanks Geva for your reply. I really apologize for responding late to your post. Actually i wanted a break from this question because i thought, may be, i am over reading this question.Geva@MasterGMAT wrote: If I may interject, what Mitch is trying to say is that the question stem allows two possible values for the probability of k=10: either 1/11 (if the set includes 10), or 0 (if the set does not include 10).
So if stat. (2) says that the probability k=10 is the same as the probability of k=-10, all we can really learn from this is that 10 and -10 are both inside the set or both outside the set together: Either p(10) = p(-10) = 1/11 (both are in the set), or p(10) = p(-10) = 0 (if both are out of the set). But we still have two possible answers as to the probability of k=10, which is why stat. (2) is insufficient.
To be honest, earlier after reading the statement 2 what i did understand:
and did not even thought about the other consecutive integers that does not include 10 {12, 14, 16...28, 30, 32} (I was also not able to understand after watching the video)the probability k=10 is the same as the probability of k=-10 means --> {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}
and this answers the question. So, answer is D
But, once again, after reading the explanation described by Mitch and reading your following explanation i have understood statement 2
the probability k=10 is the same as the probability of k=-10, all we can really learn from this is that 10 and -10 are both inside the set or both outside the set together
Thank you Mitch and Geva for helping me in this question. Really appreciate for your help.
Regards
Saurabh