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Old 03-12-2014, 07:35 PM   #1
EnoughSaid
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Default A Statistics Question For You

So my teacher gave us this stupid hard question and I cannot figure it out whatsoever. Here:

A regular six-sided die is rolled until a 5 appears. What is the probability that this process will end on an even-numbered trial? Give a reduced fraction.

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Old 03-12-2014, 07:38 PM   #2
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Default Re: A Statistics Question For You

1/6
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Old 03-12-2014, 07:40 PM   #3
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Default Re: A Statistics Question For You

The probability of rolling a 5 is obviously 1/6. But how do you determine what to do for the even-numbered trial thing?
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Old 03-12-2014, 07:47 PM   #4
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Default Re: A Statistics Question For You

Wow kids are getting dumber by the day.
Prob of getting 5 is 1/6. Prob of landing on an even trial is 1/2.
Both probabilities multiplied is 1/12.
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Old 03-12-2014, 07:52 PM   #5
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Default Re: A Statistics Question For You

Quote:
Originally Posted by Jello
Wow kids are getting dumber by the day.
Prob of getting 5 is 1/6. Prob of landing on an even trial is 1/2.
Both probabilities multiplied is 1/12.

Yeah but it says this process will end so you've already got the 5 right? When the process ends you've gotten the 5 already.Just a matter of even vs. odd trial. 1/2.

But math is my worst subject.
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Old 03-12-2014, 07:54 PM   #6
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Default Re: A Statistics Question For You

Quote:
Originally Posted by Boarder Patrol
Yeah but it says this process will end so you've already got the 5 right? When the process ends you've gotten the 5 already.Just a matter of even vs. odd trial. 1/2.

But math is my worst subject.
What? No you idiot. It's asking what is the probability of rolling a 5 on an even trial.
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Old 03-12-2014, 07:54 PM   #7
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Quote:
Originally Posted by EnoughSaid
The probability of rolling a 5 is obviously 1/6. But how do you determine what to do for the even-numbered trial thing?

Maybe if you were studying instead of at the crib chillin' you would know. Damn kid. Read a book and use Google
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Old 03-12-2014, 11:21 PM   #8
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Default Re: A Statistics Question For You

Hmm... I got 5/11.

Prob(end on an even-numbered trial)

= Prob(end on trial #2) + Prob(end on trial #4) + Prob(end on trial #6) + ...

= (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + ...

= (1/6)*[(5/6) + (5/6)^3 + (5/6)^5 + ...]

= (1/6)*(5/6)*[1 + (5/6)^2 + (5/6)^4 + ...]

The term in brackets is the sum of an infinite geometric sequence, where each successive term differs from the previous term by a factor of (5/6)^2. So, this sum is equal to 1/(1 - (5/6)^2) = 1/(1 - (25/36)) = 36/11.

So, the desired probability is (1/6)*(5/6)*(36/11) = 5/11.

Hope this helps!
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Old 03-12-2014, 11:27 PM   #9
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Default Re: A Statistics Question For You

Quote:
Originally Posted by Jello
Prob of getting 5 is 1/6. Prob of landing on an even trial is 1/2.
Both probabilities multiplied is 1/12.

If the events A = {get 5} and B = {land on an even trial} were independent events, then it would be true that Prob(A and B) = Prob(A)*Prob(B). But A and B are not independent in this case. So we can't just multiply the probabilities.
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Old 03-12-2014, 11:33 PM   #10
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Default Re: A Statistics Question For You

Quote:
Originally Posted by calculus09
Hmm... I got 5/11.

Prob(end on an even-numbered trial)

= Prob(end on trial #2) + Prob(end on trial #4) + Prob(end on trial #6) + ...

= (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + ...

= (1/6)*[(5/6) + (5/6)^3 + (5/6)^5 + ...]

= (1/6)*(5/6)*[1 + (5/6)^2 + (5/6)^4 + ...]

The term in brackets is the sum of an infinite geometric sequence, where each successive term differs from the previous term by a factor of (5/6)^2. So, this sum is equal to 1/(1 - (5/6)^2) = 1/(1 - (25/36)) = 36/11.

So, the desired probability is (1/6)*(5/6)*(36/11) = 5/11.

Hope this helps!
This inherently doesn't make sense. The question's first qualification is 1/6 of a chance of cube to land on 5, then adding a qualifier of landing 5 on an even trial. You are suggesting the probability is higher to meet both qualifications than it is to meet one.
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Old 03-12-2014, 11:45 PM   #11
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Default Re: A Statistics Question For You

Quote:
Originally Posted by calculus09
If the events A = {get 5} and B = {land on an even trial} were independent events, then it would be true that Prob(A and B) = Prob(A)*Prob(B). But A and B are not independent in this case. So we can't just multiply the probabilities.
They are independent. Whether it falls on an even roll does not affect what face is shown on the die.
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Old 03-12-2014, 11:50 PM   #12
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Default Re: A Statistics Question For You

Answer is 1/2

The fact that it has six sides means nothing.
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Old 03-12-2014, 11:56 PM   #13
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Default Re: A Statistics Question For You

Quote:
Originally Posted by shlver
They are independent. Whether it falls on an even roll does not affect what face is shown on the die.

The probability of ending on an odd-numbered trial is higher than the probability of ending on an even-number trial. The former is 6/11 (using similar reasoning as in my first post). The latter is 5/11. So, the probability of the process ending does depend on whether you consider the last trial to be even or odd. So, the events are not independent.

Last edited by calculus09 : 03-13-2014 at 12:02 AM.
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Old 03-12-2014, 11:57 PM   #14
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Default Re: A Statistics Question For You

Quote:
Originally Posted by Jello
Wow kids are getting dumber by the day.
Prob of getting 5 is 1/6. Prob of landing on an even trial is 1/2.
Both probabilities multiplied is 1/12.



This.
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Old 03-13-2014, 12:01 AM   #15
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Default Re: A Statistics Question For You

Quote:
Originally Posted by shlver
This inherently doesn't make sense. The question's first qualification is 1/6 of a chance of cube to land on 5, then adding a qualifier of landing 5 on an even trial. You are suggesting the probability is higher to meet both qualifications than it is to meet one.

We're not adding the qualifier "process ends on an even trial" to the event {land on 5 on a given roll}. We're adding the qualifier to the event {process ends}. Prob(land on 5 on a given roll) = 1/6. Prob(process ends) = Prob(first time you land on 5 is Trial 1) + Prob(first time you land on 5 is Trial 2) + Prob(first time you land on 5 is Trial 3) + ..., which isn't necessarily equal to 1/6.
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