Chapter 8.3, Problem 22PSC

To determine

To calculate: The probability that the two selected triangles are similar.

Answer to Problem 22PSC

The probability of selecting two similar triangles = $\frac{4}{15}$ .

Explanation of Solution

Given information : Here, we have given that out of six triangles, two triangles are similar.

Formula used : Probability(event) = $\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}$

Calculation: As, there are three type of triangles, three of $(30°,50°,100°)$ , two of $(40°,60°,80°)$ and one of $(60°,50°,70°)$

Probability of selecting one triangle of $(30°,50°,100°)$ from 6 triangles = $\frac{3}{6}=\frac{1}{2}$

Probability of selecting second triangle of $(30°,50°,100°)$ from remaining 5 triangles = $\frac{2}{5}$

  $\therefore$ Probability of selecting both triangles = $\frac{1}{2}\times \frac{2}{5}=\frac{1}{5}$

Now, Probability of selecting one triangle of $(40°,60°,80°)$ from 6 triangles = $\frac{2}{6}=\frac{1}{3}$

Probability of selecting second triangle of $(40°,60°,80°)$ from remaining 5 triangles = $\frac{1}{5}$

  $\therefore$ Probability of selecting both triangles = $\frac{1}{3}\times \frac{1}{5}=\frac{1}{15}$

  $\Rightarrow$ Probability of selecting two similar triangles = $\frac{1}{5}+\frac{1}{15}=\frac{3+1}{15}=\frac{4}{15}$