Answer to Question #205423 in Statistics and Probability for Farhaan

Question

A die is thrown repeatedly until a 4 and a 5 was obtained. Find the expected number of throws required?

Answer 1

Solution:

Let X denote the number throws until we get a 4. Then X takes values: $1,2,3, \ldots$ So the state space of X is $\{1,2,3, \ldots\}$

The PMF is (probability for a 4 is $\frac{1}{6}$ and for the probability of non occurrence of six is $\left(1-\frac{1}{6}\right)$ )

$f(x)=\left\{\begin{array}{cl}\left(1-\frac{1}{6}\right)^{x-1} \frac{1}{6}, & x=1,2,3, \ldots \\ 0, & \text { elsewhere }\end{array}\right.$

The expected number of throws :

$E(X)=\sum_{x=1}^{\infty} x\left(1-\frac{1}{6}\right)^{x-1} \frac{1}{6} \\=\frac{1}{6}[1+2(1-\frac{1}{6})+3(1-\frac{1}{6})^2+4(1-\frac{1}{6})^3+...] \\=\frac{1}{6}(1-\frac{5}{6})^{-2} \\=\frac{1}{6}(\frac{1}{6})^{-2} \\=6$

Similarly, to get a 5, the expected number of throws is 6.

Thus, in total, the expected number of throws to get a 4 and a 5 is (6+6)=12.