Chapter 5, Page 197, Ques 7

A die is rolled until the first time T that a six turns up.

  1. What is the probability distribution for T?

Per geometric distribution P(T = j) = \({q}^{j-1}\)\(\times{p}\) where q=1-p

Here p = 1/6 so q = 1-1/6 = 5/6

So probability distribution for T is P(T=j) = \({(\frac{5}{6})}^{j-1}\)\(\times{\frac{1}{6}}\)

for j = 1, 2, 3, …..

# draw curve for probability distribution function
curve( ((5/6)^(x-1))*(1/6), from = 1, to = 100)

  1. Find P(T > 3).

Using the formula

P(T > k) = \(\sum\limits_{i=1}^n {q}^{j-1}p\) = \({q}^k\)

so for P(T > 3) = \({q}^3\) = \(({\frac{5}{6}})^3\) = 125/216 = 0.59

  1. Find P(T > 6|T > 3).

Using the formula P(T >r + s | T > s) = \({q}^s\)

so for P(T > 6|T > 3) = \(({\frac{5}{6}})^3\) = 125/216 = 0.59