Chapter 1.9, Problem 9PSB

To determine

The probability that an acute angle is selected both times.

Answer to Problem 9PSB

  $Probabilitythatbothanglesareacute= \frac{9}{10}$

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:

Total number of ways of selecting 2 angles out of 5 = $C_2^5=\frac{5!}{2!\left(5-2\right)!}=\frac{5!}{2!\left(3\right)!}=\frac{120}{2\times 6}=10$

  $Numberofwaysofgetting2acuteangleafterreplacement=C_1^3\times C_1^3=\frac{3!}{1!\left(3-1\right)!}\times \frac{3!}{1!\left(3-1\right)!}=3\times 3=9$

  $\begin{array}{l}Probabilityofgettingbothangleacute=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}=\frac{9}{10}\\ \end{array}$