Chapter 5.6, Problem 21PSD

a.

To determine

To calculate:Theprobability that the angles are congruent if two of the eight angles are selected at random.

Answer to Problem 21PSD

Theprobabilityis $\frac{2}{7}$ .

Explanation of Solution

Given information:

The angles of a rectangle and parallelogram that is not a rectangle are in a box.

Formula used:

Probability of event to happen is

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

Calculation:

  

  

The probability that the angles are congruent if two of the eight angles are selected at random is $=\frac{Correctangles}{Possibleangles}=\frac{8}{28}=\frac{2}{7}$

b.

To determine

To find:Whether to take man’s bet of $\text{\$}20$ .

Answer to Problem 21PSD

No, you should not take man’s bet.

Explanation of Solution

Given information:

The angles of a rectangle and parallelogram that is not a rectangle are in a box.

A man offers to have two tries at getting a pair of congruent angles. In other words, you would draw a pairs of angles at random, then replace the pair, and then draw a pair again. The man is willing to bet you $\text{\$}20$ that you won’t draw a congruent pair either time.

Formula used:

Probability of event to happen is

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

  

  

The probability that the angles are congruent if two of the eight angles are selected at random is $p=\frac{Correctangles}{Possibleangles}=\frac{8}{28}=\frac{2}{7}$

The complementary of p is the p of draws that won’t win, i.e. 1 - p.

  $1-p=1-\frac{2}{7}=\frac{5}{7}$

The p that is neither a winner is $\frac{5}{7}\times \frac{5}{7}=\frac{25}{49}$

The p of winning is $1-\frac{25}{49}=\frac{24}{49}$

No, you should not take man’s bet because you will win $\frac{24}{49}$ of the time and lose $\frac{25}{49}$ of

the time.