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.
:wtf: :wtf:
Re: A Statistics Question For You
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? :banghead:
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.
Re: A Statistics Question For You
[QUOTE=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.[/QUOTE]
Yeah but it says this [I]process will end[/I] 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.
Re: A Statistics Question For You
[QUOTE=Boarder Patrol]Yeah but it says this [I]process will end[/I] 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.[/QUOTE]
What? No you idiot. It's asking what is the probability of rolling a 5 on an even trial.
Re: A Statistics Question For You
[QUOTE=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? :banghead:[/QUOTE]
Maybe if you were studying instead of at the crib chillin' you would know. Damn kid. Read a book and use Google
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!
Re: A Statistics Question For You
[QUOTE=Jello]
Prob of getting 5 is 1/6. Prob of landing on an even trial is 1/2.
Both probabilities multiplied is 1/12.[/QUOTE]
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.
Re: A Statistics Question For You
[QUOTE=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![/QUOTE]
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.
Re: A Statistics Question For You
[QUOTE=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.[/QUOTE]
They are independent. Whether it falls on an even roll does not affect what face is shown on the die.
Re: A Statistics Question For You
Answer is 1/2
The fact that it has six sides means nothing.
Re: A Statistics Question For You
[QUOTE=shlver]They are independent. Whether it falls on an even roll does not affect what face is shown on the die.[/QUOTE]
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.
Re: A Statistics Question For You
[QUOTE=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.[/QUOTE]
:oldlol:
This.
Re: A Statistics Question For You
[QUOTE=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.[/QUOTE]
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.