Q.7.4 - A First Course in Probability

Question

If a die is to be rolled until all sides have appeared at least once, find the expected number of times that outcome $1$ appears.

Step-by-Step Solution

Answer

The expected number of times that outcome $1$ appears is 2.45

Step 1: Given Information

 If a die is to be rolled until all sides have appeared at least once,find the expected number of times that outcome $1$ appears.

Step 2: Explanation

It is known that if we roll a six-sided fair die, there are $6$ possible outcomes, each one with a probability value $p=\frac{1}{6}$. Assume that die is rolled until all sides have appeared at least once, and in that case, let X be the number of rolls. Further, let random variable Xi represents the number of rolls needed to arrive at a new face on the die if i different faces have already appeared up. Then,

$X=1+X_1+X_2+X_3+X_4+X_5$

$p_2 = \frac{4}{6}$, X$3$ is a geometric random variable with parameters $p_{3 }= \frac{3}{6}$, X$4$ is a geometric random variable with parameters $p_4 = \frac{2}{6}$ and X5 is a geometric random variable with parameters $p_{5 }= \frac{1}{6}$. Therefore,

$E\left[X\right]=1+\sum _{i=1}^{5}E\left[X_i\right]=1+\sum _{i=1}^5\frac{1}{p_i}=1+(\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1})=14.7.$

Step 3: Explanation

Now, let's define indicator variables  Ij, j=$1,2$, ...., X as:

$I_j=\{\begin{array}{ll}1,& \text{if}{E}_1\text{occurs}\\ 0,& \text{if}{E}_1\text{does not occur}\end{array}$

Whereby $E_1$ denote the event :

$E_1="\text{the outcome}1\text{appears " ,}P\left\{E_1\right\}=p\text{.}$

Then, if Y  represents the number of times that outcome $1$ appears, we have that

$Y=\sum _{j=1}^X{I}_j$

and therefore the expected number of time that outcome$1$ appear is:

$E\left[Y\right]=E\left[\sum _{j=1}^X{I}_j\right]=E\left[E[\sum _{j=1}^X{I}_j\mid X]\right]=E\left[X\underset{=p}{\underbrace{P\left\{E_1\right\}}}\right]=pE\left[X\right]=\mathbf{2}\mathbf{.}\mathbf{45}.$

Step 4: Final Answer

The expected number of times that outcome $1$ appears is 2.45