Chapter 13.3, Problem 30AYU

Assigning Probabilities A die is weighted so that a six cannot appear. All the other faces occur with the same probability. What probability should be assigned to each face?

To determine

What probability should be assigned to each face?

Answer to Problem 30AYU

$P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=1/5$, $P\left(6\right)=0$

Explanation of Solution

Given:

A die is weighted so that a six cannot appear. All the other faces occur with the same probability.

Formula used:

${\displaystyle \sum }_{i=1}^{n}P\left(e_i\right)=P\left(e_1\right)+P\left(e_2\right)+\cdot \cdot \cdot +P\left(e_n\right)=1$

Calculation:

A die is weighted so that a six cannot appear. All the other faces occur with the same probability.

Outcomes of die are $\text{1, 2, 3, 4, 5, 6}$.

Let $P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=x$, $P\left(6\right)=0$

$x+x+x+x+x=1$

$x=\frac{1}{5}$

$P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=1/5$, $P\left(6\right)=0$