Question

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

Short answer

The probability is$P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=\frac{1}{5}$.

Step-by-step solution

Step 1. Given information. 

A die is weighted so that a six cannot appear. The other faces occur with the same probability. What probability should we assign to each face?

Step 2. Find probability.  

We are given that, the six cannot appear and the other numbers in a die has the same probability. Since the sample space of a die is

$S=\left\{1,2,3,4,5\right\}$

Thus, we can let

$$\begin{gathered} P\left(6\right)=0 \\ P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=x \end{gathered}$$

Since the sum of the probabilities of the possible outcomes must equal 1.

$P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)=1$

Substituting the given, and solving for $x$, we have

$$\begin{gathered} P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)=1 \\ x+x+x+x+x+0=1 \\ 5x=1 \\ x=\frac{1}{5} \end{gathered}$$

Step 3. Substitute the value. 

Since $x=\frac{1}{5}$assign the probabilities

$P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=x=\frac{1}{5}$.

Step 4. Conclusion.

The probability is$P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=\frac{1}{5}$