Chapter 9.7, Problem 28PSD

a.

To determine

To calculate: The probability that two triangles are the members of the $3-4-5$ family.

Answer to Problem 28PSD

The probability is $\frac{5}{14}$ .

Explanation of Solution

Given information:

A bag contains eight right triangles. Five are the members of $3-4-5$ family, and two are ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangles. A puppy falls over the bag, and two triangles fall out on the floor.

Formula used:

Probability of event to happen is

  $P\left(E\right)=\frac{Numberoffavourableoutcomes}{TotalNumberofoutcomes}$

Calculation:

The tree diagram is as follows:

  

The probability that two triangles are the members of the $3-4-5$ family is $=\frac{5}{8}\times \frac{4}{7}=\frac{5}{14}$

b.

To determine

To find: The probability that at least one of the triangles is the members of the $3-4-5$ family.

Answer to Problem 28PSD

The probability is $\frac{25}{28}$ .

Explanation of Solution

Given information:

A bag contains eight right triangles. Five are the members of $3-4-5$ family, and two are ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangles. A puppy falls over the bag, and two triangles fall out on the floor.

Formula used:

Probability of event to happen is

  $P\left(E\right)=\frac{Numberoffavourableoutcomes}{TotalNumberofoutcomes}$

Calculation:

The tree diagram is as follows:

  

The probability that at least one of the triangles is the members of the $3-4-5$ family is

  $\begin{array}{l}=\left(\frac{5}{8}\times \frac{4}{7}\right)+\left(\frac{5}{8}\times \frac{2}{7}\right)+\left(\frac{5}{8}\times \frac{1}{7}\right)+\left(\frac{2}{8}\times \frac{5}{7}\right)+\left(\frac{1}{8}\times \frac{5}{7}\right)\\ =\frac{10}{28}+\frac{5}{28}+\frac{5}{56}+\frac{5}{28}+\frac{5}{56}\\ =\frac{20}{28}+\frac{10}{56}\\ =\frac{20}{28}+\frac{5}{28}\\ =\frac{25}{28}\end{array}$

c.

To determine

To find: The probability that one of the triangle is the members of the $3-4-5$ family and other triangle is ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ .

Answer to Problem 28PSD

The probability is $\frac{5}{14}$ .

Explanation of Solution

Given information:

A bag contains eight right triangles. Five are the members of $3-4-5$ family, and two are ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangles. A puppy falls over the bag, and two triangles fall out on the floor.

Formula used:

Probability of event to happen is

  $P\left(E\right)=\frac{Numberoffavourableoutcomes}{TotalNumberofoutcomes}$

Calculation:

The tree diagram is as follows:

  

The probability that one of the triangle is the members of the $3-4-5$ family and other is ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangle is

  $\begin{array}{l}=\left(\frac{5}{8}\times \frac{2}{7}\right)+\left(\frac{2}{8}\times \frac{5}{7}\right)\\ =\frac{5}{28}+\frac{5}{28}\\ =\frac{10}{28}\\ =\frac{5}{14}\end{array}$