A Statistics Question For You

[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 :

[Myth]

Holy shit you guys.

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