The probability that one gets a sum of 8 in the first four tosses of the dice is
.
Thus what is the probability that it takes more than four tosses?
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?
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 ?