Chapter 5 - Exercise 7, Page 197

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

(a) What is the probability distribution for \(T\)?

We are going to be rolling the dice until we have a 6, which has a \(\frac{1}{6}\) chance of occuring.

We also know from the textbook the formula for solving this is \(P(T = n) = q^{n-1}\times p\) where \(q = 1-p\).

\(P(n) = q^{n-1}\times p\)

\(P(n) = (1-\frac{1}{6})^{n-1}\times\frac{1}{6}\)

\(P(n) = (\frac{5}{6})^{n-1}\times\frac{1}{6}\)

So the probability distribution for T is \(P(n) = (\frac{5}{6})^{n-1}\times\frac{1}{6}\) where \(n\) is the number of trials performed, ranging from 0 to infinity.

curve((1/6)*((5/6)^(x-1)), from = 0, to=50)

(b) Find \(P(T>3)\).

From the text book, we also know how to solve this. When T is greater than a number that is given, the formula can be reduced to \(P(T > k) = q^{k}\).

\(P(T > k) = q^{k}\)

\(P(T > 3) = (1-p)^{3}\)

\(P(T > 3) = (1-\frac{1}{6})^{3}\)

\(P(T > 3) = \frac{5}{6}^{3}\)

\(P(T > 3) = \frac{125}{216}\)

(c) Find \(P(T>6|T>3)\).

Again, the textbook provides a formula that shortens this down quite a bit. \(P(T>r+s|T>s) = q^{s}\).

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

\(P(T>6|T>3) = (1-p)^{3}\)

\(P(T>6|T>3) = (1-\frac{1}{6})^{3}\)

\(P(T>6|T>3) = \frac{5}{6}^{3}\)

\(P(T>6|T>3) = \frac{125}{216}\)