Chapter 1.9, Problem 7PSB

To determine

The probability that one of the selected angle is obtuse and other acute.

Answer to Problem 7PSB

  $Probabilitythatoneoftheangleisobtuse= \frac{3}{10}$

Explanation of Solution

Given:

The five angles are given

Number of acute angle ( ${50}^{\circ },{60}^{\circ },{40}^{\circ }$ ) = 3

Number of obtuse angle ( ${130}^{\circ }$ ) = 1

Number of right angle ( ${90}^{\circ }$ ) = 1

Formula used:

  $\begin{array}{l}Probabilityofaevent=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}\\ \end{array}$

  $C_r^n=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Obtuse angle is the angle whose value is more than ${90}^{\circ }$

Total number of ways of selecting 2 angles out of 5

= $C_2^5=\frac{5!}{2!\left(5-2\right)!}=\frac{5!}{2!\left(3\right)!}=\frac{120}{2\times 6}=10$

Number of ways of selecting 1 acute and 1 obtuse angle

= $C_1^3\times C_1^1=\frac{3!}{1!\left(3-1\right)!}\times \frac{1!}{1!\left(1-1\right)!}=\frac{6}{2}\times \frac{1}{1}=3$

  $\begin{array}{l}Probabilityofagettinganobtuseangle=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}=\frac{3}{10}\\ \end{array}$