Chapter 13.3, Problem 29AYU

Assigning Probabilities A die is weighted so that an odd-numbered face is twice as likely to occur as an even-numbered face. What probability should be assigned to each face?

To determine

What probability should be assigned to heads? To tails?

Answer to Problem 29AYU

$P\left(2\right)=P\left(4\right)=P\left(6\right)=1/9$

$P\left(1\right)=P\left(3\right)=P\left(5\right)=2/9$

Explanation of Solution

Given:

A die is weighted so that an odd-numbered face is twice as likely to occur as an even-numbered face.

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 an odd-numbered face is twice as likely to occur as an even-numbered face.

Outcomes of die are $1,2,3,4,5,6$.

Let $P\left(2\right)=P\left(4\right)=P\left(6\right)=x$,

$P\left(1\right)=P\left(3\right)=P\left(5\right)=2x$

Sum of the probability of individual event is 1.

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

$x+x+x+2x+2x+2x=1$

$x=\frac{1}{9}$

$P\left(2\right)=P\left(4\right)=P\left(6\right)=1/9$

$P\left(1\right)=P\left(3\right)=P\left(5\right)=2/9$