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.

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.

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.

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.

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.

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.

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.

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.

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.