Chapter 0.11, Problem 10E

To determine

The probability of the given data.

Answer to Problem 10E

  $\mathrm{P}\left(morethan2rightangles\right)=\frac{2}{6}=\frac{1}{3}$

Explanation of Solution

Given :

  $\begin{array}{l}triangle=3\\ square=2\\ pentagon=1\end{array}$

Solution:

A square 4 right-angles.

  $\begin{array}{l}totaloutcome=3+2+1\\ totaloutcome=6\end{array}$

Probability:

  $\mathrm{Pr}obability=\frac{probableoutcome}{totaloutcome}$

  $\begin{array}{l}probableoutcome=2\\ \mathrm{P}\left(morethan2rightangles\right)=\frac{2}{6}=\frac{1}{3}\end{array}$