Question

A loaded 8-sided die is loaded so that the number 4 occurs 3/10 of the time while the other numbers occur with equal frequency. what is the expected value of this die

Answer 1

Answer:

4.4

Step-by-step explanation:

The sum of the probabilities of all possible outcomes is 1.

As the die is loaded so that the number 4 occurs 3/10 of the time, and the other numbers occur with equal frequency, then the probability of numbers 1, 2, 3, 5, 6, 7 and 8 being rolled is 1/10.

Create a probability distribution table for X, where X is the score on the loaded 8-sided die:

$\begin{array}{ | c | c | c | c | c | c | c | c | c |}\cline{1-9}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\cline{1-9} & & & & & & & &\\ \text{P}(X=x) & \dfrac{1}{10} & \dfrac{1}{10} & \dfrac{1}{10} & \dfrac{3}{10} & \dfrac{1}{10} & \dfrac{1}{10} & \dfrac{1}{10} & \dfrac{1}{10}\\& & & & & & & &\\\cline{1-9}\end{array}$

Add a product row and a totals column:

$\begin{array}{ | c | c | c | c | c | c | c | c | c | c |}\cline{1-10}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &\text{Total} \\ \cline{1-10} &&&&&&&&& \\ \text{P}(X=x) & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & 1\\ &&&&&&&&&\\ \cline{1-10} &&&&&&&&&\\\text{Product} & \frac{1}{10} & \frac{2}{10} & \frac{3}{10} & \frac{12}{10} & \frac{5}{10} & \frac{6}{10} & \frac{7}{10} & \frac{8}{10} & \frac{44}{10} \\ &&&&&&&&&\\\cline{1-10}\end{array}$

(The Product is row is the product of the score on the die and its probability).

The expected value (EV) is the sum of the product of each outcome and its probability.  

Therefore, the expected value (EV) of this die is 44/10 = 4.4