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 03-12-2014, 08:35 PM #1 EnoughSaid Gooner | Heat Nation     Join Date: Jul 2010 Posts: 6,499 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.
 03-12-2014, 08:38 PM #2 CelticBaller Killer Instinct     Join Date: May 2010 Location: CT, GOATLAND Posts: 21,325 Re: A Statistics Question For You 1/6
 03-12-2014, 08:40 PM #3 EnoughSaid Gooner | Heat Nation     Join Date: Jul 2010 Posts: 6,499 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?
 03-12-2014, 08:47 PM #4 Jello Hume>Kunt.   Join Date: Nov 2009 Posts: 1,531 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.
03-12-2014, 08:52 PM   #5
Boarder Patrol
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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.

03-12-2014, 08:54 PM   #6
Jello
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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.

03-12-2014, 08:54 PM   #7
D-FENS
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Re: A Statistics Question For You

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

 03-13-2014, 12:21 AM #8 calculus09 Saw a basketball once   Join Date: Feb 2011 Posts: 6 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!
03-13-2014, 12:27 AM   #9
calculus09

Join Date: Feb 2011
Posts: 6
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.

03-13-2014, 12:33 AM   #10
shlver
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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.

03-13-2014, 12:45 AM   #11
shlver
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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.

 03-13-2014, 12:50 AM #12 ace23 *****     Join Date: May 2011 Location: Houston, TX Posts: 9,487 Re: A Statistics Question For You Answer is 1/2 The fact that it has six sides means nothing.
03-13-2014, 12:56 AM   #13
calculus09

Join Date: Feb 2011
Posts: 6
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 01:02 AM.

03-13-2014, 12:57 AM   #14
YouGotServed

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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.

03-13-2014, 01:01 AM   #15
calculus09

Join Date: Feb 2011
Posts: 6
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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