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BTG practice Question - 700 level (probability)

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BTG practice Question - 700 level (probability)

by saurabhkamal1981 » Tue Mar 08, 2011 3:39 am
*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
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by GMATGuruNY » Tue Mar 08, 2011 4:01 am
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 in the set, P(k=10) = 1/11.
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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by saurabhkamal1981 » Tue Mar 08, 2011 5:52 am
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.
Hi Mitch,

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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by GMATGuruNY » Tue Mar 08, 2011 10:23 am
saurabhkamal1981 wrote:
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.
Hi Mitch,

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 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?
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by saurabhkamal1981 » Wed Mar 09, 2011 3:52 am
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?
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.

Regards
Saurabh
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by Geva@EconomistGMAT » Wed Mar 09, 2011 4:14 am
saurabhkamal1981 wrote:
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?
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.

Regards
Saurabh
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.
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by saurabhkamal1981 » Thu Mar 10, 2011 7:11 am
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.
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.

To be honest, earlier after reading the statement 2 what i did understand:
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
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)

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
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