Chapter 10, Problem 2RE

Solution

To calculate: The probabilities of the three outcomes.

  $P\left(A\right)=\frac{3}{10},P\left(B\right)=\frac{6}{10},P\left(C\right)=\frac{1}{10}.$

Given information:

Sample space is $\left\{A,B,C\right\}$ .

Calculation:

Now suppose x is the probability outcome of A .

So,

  $P\left(A\right)=x$

Since, it is given that probability of A is half likely to occur to B .

So,

  $\begin{array}{c}P\left(B\right)=2P\left(A\right)\\ =2x\end{array}$

And similarly, it is given that probability of A is three times likely to occur to C .

So,

  $\begin{array}{c}P\left(C\right)=\frac{P\left(A\right)}{3}\\ =\frac{x}{3}\end{array}$

As it is known that for valid probabilities sum of all probabilities must be equal to 1.

So,

  $\begin{array}{c}P\left(A\right)+P\left(B\right)+P\left(C\right)=1\\ x+2x+\frac{x}{3}=1\\ \frac{10x}{3}=1\\ x=\frac{3}{10}\end{array}$

Now, Replace $x$ by $\frac{3}{10}$ in the probabilities of the three outcomes:

  $\begin{array}{l}P\left(A\right)=x=\frac{3}{10}\\ P\left(B\right)=2x=\frac{6}{10}\\ P\left(C\right)=\frac{x}{3}=\frac{1}{10}.\end{array}$

Thus, the probabilities of three outcomes are $P\left(A\right)=\frac{3}{10},P\left(B\right)=\frac{6}{10},P\left(C\right)=\frac{1}{10}.$