Grimstead, p. 197
7. A die is rolled until the first time T that a six turns up.
(a) What is the probability distribution for T?
This is the geometric distribution where P(X=k) = p(1−p)k for k ≥ 0, p=1/6
We’re looking for the n number of trials it will take to get a success, given probability p of success.
Examples of p=1/6 and p=0.5 for comparison:
par(mfrow=(c(1,2)))
plot(pgeom(0:20, prob=1/6), type='h')
plot(pgeom(0:20, prob=.5), type='h')
The probability of getting a success grows quickly when p is fair and more slowly when p is smaller.
(b) Find P (T > 3).
We find the complement of three successive failures.
p = .1666
P(T > 3) = 1-(.166)4-1x.166
= 0.8843
Using R’s distribution function:
round(1-dgeom(2, prob=1/6),4)## [1] 0.8843(c) P (T>6 | T>3)
In the geometric distribution, the mean number of trials before a success is 1/p, or 6 in this dice-rolling example. So in 6 rolls we should get one six.
Each roll of the die is independent, so P(A|B) = P(A) and vice versa.
So P(T>6 | T>3) = P(T>6) = 1-(.166)7-1x.166 = 0.9442