What is the probability of one dice rolled twice that will result in exactly one three being rolled?

Also what is the probability of at least one three being rolled (one dice rolled twice)? I have the answers, I just need to understand it…

The answer for exactly one three comes up is 5/18 and at least one three comes up is 11/36. How did they get 5/18?

*Question posted by: atukomeyer*

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September 24th, 2009 at 3:56 pm

Kansieo.com

p=probability

In this case it is 1/6

The chance of getting ONE 3 after 2 rolls is:

((1-p)*p)*2

(5/6*1/6)*2=5/18=0.625

The chance of getting ONE 3 after n rolls is:

((1-p)^(n-1)*p)*n

—————————–

The chance of getting at least one 3 is:

p+(1-p)*p

1/6+5/6*1/6=11/36=0.30555

The chance of doing it in 3 turns is 1/6+5/6*1/6+5/6²*1/6

The chance of doing it within n turns is

n

?(1-p)^(x-1)*p

x=1

or

1-(1-p)^n

I realised that I made a small mistake. I have corrected it above.

I was doing the calculation for doing it on the nth turn instead of WITHIN n turns. For instance there is a 5/6 chance of not getting a 3 on your first roll and a 1/6 chance to get a 3 on your second roll. There is also a 1/6 chance of getting a 3 on your first roll and a 5/6 chance of not getting a 3 on your second roll.(5/6*1/6)+(1/6*5/6).

I hope that clears things up for you.