91Ó°ÊÓ

A die is weighted so that an odd-numbered face is twice as likely to occur as an even numbered face. What probability should we assign to each face?

We are given that the odd numbered face Ois twice as likely to occur as an even numbered face E in a die. Thus, we can let

$$\begin{gathered} P\left(\mathrm{E}\right)=x \\ P\left(\mathrm{O}\right)=2x \end{gathered}$$

Considering the sample space of a die

$S=\left\{1,2,3,4,5,6\right\}$

We can notice that there are 3 odd and 3 even numbers. Thus our equation will be

$3×P\left(\mathrm{E}\right)+3×P\left(\mathrm{O}\right)=1$

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

Substituting the value of $P\left(\mathrm{E}\right)andP\left(\mathrm{O}\right)$and solving for $x$, we have

localid="1647256783031" $$\begin{gathered} 3×P\left(\mathrm{E}\right)+3×P\left(\mathrm{O}\right)=1 \\ 3×\left(x\right)+3×\left(2x\right)=1 \\ 3x+6x=1 \\ 9x=1 \\ x=\frac{1}{9} \end{gathered}$$

Since $x=\frac{1}{9}$, assign the probabilities

$P\left(\mathrm{E}\right)=x=\frac{1}{9},P\left(\mathrm{O}\right)=2x=\frac{2}{9}$

Since $x=\frac{1}{9}$assign the probabilities

$P\left(\mathrm{E}\right)=x=\frac{1}{9},P\left(\mathrm{O}\right)=2x=\frac{2}{9}$

Thus, the probability of each number face is

localid="1647259369578" $$\begin{gathered} P\left(\mathrm{O}\right)=P\left(1\right)=P\left(3\right)=P\left(5\right)=\frac{2}{9} \\ P\left(\mathrm{E}\right)=P\left(2\right)=P\left(4\right)=P\left(6\right)=\frac{1}{9} \end{gathered}$$

The probability is$P\left(1\right)+P\left(3\right)+P\left(5\right)=\frac{2}{9};P\left(2\right)+P\left(4\right)+P\left(6\right)=\frac{1}{9}$.