# Random Variable Help

• Jun 19th 2007, 03:08 PM
engineer22
Random Variable Help
A pair of fair dice are tossed until a sum of eight appears for the first time. What is the probability that more than four rolls will be required for that to happen?

I figured that the probability that the sum is 8 is 5/36 on one throw from the choices {(2,6)(3,5)(4,4)(5,3)(6,2)} I'm not sure how to find the probability that more than 4 rolls are needed to get the sum 8 though. Can someone help?
• Jun 19th 2007, 03:24 PM
Plato
The probability that one gets a sum of 8 in the first four tosses of the dice is
$\displaystyle \sum\limits_{k = 1}^4 {\left( {\frac{5}{{36}}} \right)\left( {\frac{{31}}{{36}}} \right)^{k - 1} }$.

Thus what is the probability that it takes more than four tosses?
• Jun 19th 2007, 03:34 PM
engineer22
Would it be...

1 - http://www.mathhelpforum.com/math-he...3674d006-1.gif.

answer = 1 - 0.1388888 + 0.1195987 + .1029878 + 0.0886839 = 0.5498 ?
• Jun 19th 2007, 03:36 PM
Plato
Yes, you have the idea.
But check the calulation!
• Jun 19th 2007, 03:48 PM
engineer22
If I do it this way:

1 - 0.1388888 + 0.1195987 + .1029878 + 0.0886839

I get 1.1724 and the probability can't be more than 1

So would the calculation be correct if I did this:

( 0.1388888 + 0.1195987 + .1029878 + 0.0886839 ) = 0.4502

1 - 0.4502 = 0.5498 ?
• Jun 19th 2007, 08:34 PM
CaptainBlack
Quote:

Originally Posted by engineer22
A pair of fair dice are tossed until a sum of eight appears for the first time. What is the probability that more than four rolls will be required for that to happen?

I figured that the probability that the sum is 8 is 5/36 on one throw from the choices {(2,6)(3,5)(4,4)(5,3)(6,2)} I'm not sure how to find the probability that more than 4 rolls are needed to get the sum 8 though. Can someone help?

Alternative method:

To take more than 4 rolls requires that rolls 1 to 4 all do not show a sum of 8.

The probability of this is:

$\displaystyle (1-p(sum=8))^4 = (1-5/36)^4 \approx 0.5498$

RonL