Chapter 1.9, Problem 6PSB

To determine

The probability that both the selected angles are acute.

Answer to Problem 6PSB

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

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:

Acute angle are the angles whose value is less than ${90}^{\circ }$

∴ Number of acute angle out of 5 given angle are 3

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 2 acute angles out of 3 = $C_2^3=\frac{3!}{2!\left(3-2\right)!}=\frac{3!}{2!\left(1\right)!}=\frac{6}{2\times 1}=3$

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