A Statistics Question For You

[EnoughSaid]

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:

[CelticBaller]

1/6

[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:

[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.

[Boarder Patrol]

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.

[Jello]

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.

[D-FENS]

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:

Maybe if you were studying instead of at the crib chillin' you would know. Damn kid. Read a book and use Google

[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!

[calculus09]

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.

[shlver]

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.

[shlver]

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.

[ace23]

Answer is 1/2

The fact that it has six sides means nothing.

[calculus09]

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.

[YouGotServed]

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.

:oldlol:

This.

[calculus09]

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.

[ace23]

What? No you idiot. It's asking what is the probability of rolling a 5 on an even trial.
Lol you're wrong.

[AintNoSunshine]

1/2



Or, you can do the trial yourself and video tape it as proof so you can show the b!tch when she questions whatever answer you got

[shlver]

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 reason 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.
Well you didn't post a reason or any elaboration on your mathematics. This promotes copy and pasting and not learning. I need reference to what equation you're trying to use to understand your answer in context. If you are providing an answer you should be explaining why you used that equation.
I should have asked for this first, but the answer simply did not make sense from an intuitive standpoint.

[ace23]

People are not understanding the question. If the dye is rolled until a 5 appears, then the probability of getting a 5 does not need to be taken into account.

It's 1/2.

It'd be different if they asked for the probability of getting a 5 on an even-numbered trial.

[YouGotServed]

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!

lol


People are not understanding the question. If the dye is rolled until a 5 appears, then the probability of getting a 5 does not need to be taken into account.

It's 1/2.

It'd be different if they asked for the probability of getting a 5 on an even-numbered trial.

lol

[ace23]

The six-side shit is there to confuse you.

[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.
I understand why this reasoning is wrong. Ace23's post made me realize it.

[shlver]

People are not understanding the question. If the dye is rolled until a 5 appears, then the probability of getting a 5 does not need to be taken into account.

It's 1/2.

It'd be different if they asked for the probability of getting a 5 on an even-numbered trial.
This explained it semantically, but calculus09's answers are pretty close to what you got. I hope calculus09 provides an explanation to his mathematical justification.

[calculus09]

This explained it semantically, but calculus09's answers are pretty close to what you got. I hope calculus09 provides an explanation to his mathematical justification.

Oh. Sorry. Let me try to explain a bit more.

In the original problem, we want to find the probability of the process ending on an even-numbered trial. The event {process ends on an even-numbered trial} is the union of disjoint events of the form {process ends on Trial 2T}, where T = 1,2,3,4,...

In probability, when two events A and B are disjoint, the probability of the union of the events is equal to the sum of the probabilities. In other words, Prob(A or B) = Prob(A) + Prob(B).

So, Prob(process ends on an even-numbered trial) = Prob(process ends on Trial 2) + Prob(process ends on Trial 4) + Prob(process ends on Trial 6) + ...

Now, the question is how to find Prob(process ends on Trial 2T) for any T. Well, in order for the process to end on Trial 2T, it means that the cube lands on 5 on trial 2T and lands on something other than 5 on each of the previous 2T - 1 trials. Since the outcomes of the rolls are independent, we can multiply these 2T probabilities to get the following expression:

Prob(process ends on Trial 2T) = Prob(cube lands on 1,2,3,4 or 6 on Trial 1)*Prob(cube lands on 1,2,3,4 or 6 on Trial 2)* ... *Prob(cube lands on 1,2,3,4 or 6 on Trial 2T-1)*Prob(cube lands on 5 on Trial 2T).

Since each side of the cube has 1/6 probability of being face-up, we have the following:

Prob(cube lands on 1,2,3,4 or 6) = (1/6) + (1/6) + (1/6) + (1/6) + (1/6) = 5/6.

Prob(cube lands on 5) = 1/6.

So Prob(process ends on Trial 2T) = (5/6)^(2T-1)*(1/6).

Thus, we get the following:

Prob(process ends on an even-numbered trial)
= Prob(process ends on Trial 2) + Prob(process ends on Trial 4) + Prob(process ends on Trial 6) + ...
= (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + ...

And then the math is the same as before.

[shlver]

Oh. Sorry. Let me try to explain a bit more.

In the original problem, we want to find the probability of the process ending on an even-numbered trial. The event {process ends on an even-numbered trial} is the union of disjoint events of the form {process ends on Trial 2T}, where T = 1,2,3,4,...

In probability, when two events A and B are disjoint, the probability of the union of the events is equal to the sum of the probabilities. In other words, Prob(A or B) = Prob(A) + Prob(B).

So, Prob(process ends on an even-numbered trial) = Prob(process ends on Trial 2) + Prob(process ends on Trial 4) + Prob(process ends on Trial 6) + ...

Now, the question is how to find Prob(process ends on Trial 2T) for any T. Well, in order for the process to end on Trial 2T, it means that the cube lands on 5 on trial 2T and lands on something other than 5 on each of the previous 2T - 1 trials. Since the outcomes of the rolls are independent, we can multiply these 2T probabilities to get the following expression:

Prob(process ends on Trial 2T) = Prob(cube lands on 1,2,3,4 or 6 on Trial 1)*Prob(cube lands on 1,2,3,4 or 6 on Trial 2)* ... *Prob(cube lands on 1,2,3,4 or 6 on Trial 2T-1)*Prob(cube lands on 5 on Trial 2T).

Since each side of the cube has 1/6 probability of being face-up, we have the following:

Prob(cube lands on 1,2,3,4 or 6) = (1/6) + (1/6) + (1/6) + (1/6) + (1/6) = 5/6.

Prob(cube lands on 5) = 1/6.

So Prob(process ends on Trial 2T) = (5/6)^(2T-1)*(1/6).

Thus, we get the following:

Prob(process ends on an even-numbered trial)
= Prob(process ends on Trial 2) + Prob(process ends on Trial 4) + Prob(process ends on Trial 6) + ...
= (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + ...

And then the math is the same as before.
Wow, very well explained. It made me understand perfectly. Future repped.

[calculus09]

Wow, very well explained. It made me understand perfectly. Future repped.

You're welcome. :)

[russwest0]

Holy shit a bunch of retards in this generation.

1/2.

[riseagainst]

1/2 is the answer. The 5 is there to distract you.

[Theoo]

hey i m expert in excel and i did these for you i hope it more clear now :

http://i.imgur.com/j7258IX.png

[Myth]

Holy shit you guys. :oldlol:

The story problem trolled you guys. It is 1/2.

[DeuceWallaces]

Getting trolled on probability?

[no pun intended]

You're welcome. :)
did no one notice how this calculus09 account was made in feb 2011, yet his first posts are in this thread regarding mathematics lmfao

[Myth]

Damn, I just realized my over simplified answer of 1/2 is wrong. It is close to one half, but I didn't factor in the disadvantage evens have due to it allows following the odd attempt.

[ace23]

Damn, I just realized my over simplified answer of 1/2 is wrong. It is close to one half, but I didn't factor in the disadvantage evens have due to it allows following the odd attempt.
What?

[Boarder Patrol]

Holy shit you guys. :oldlol:

The story problem trolled you guys. It is 1/2.

Said this in the first 3 responses.

[STATUTORY]

Lol it's not one half