Chapter 6, Problem 1P

A local hardware store has a “Savings Wheel” at the checkout. Customers get to spin the wheel and. when the wheel stops, a pointer indicates how much they will save. The wheel can stop in any one of 50 sections. Of the sections, 10 produce 0% off, 20 sections are for 10%off, 10 sections for 20%, 5 for 30%, 3 for 40%, 1 for 50%, and 1 for 100% off. Assuming that all 50 sections are equally likely,

1. a. What is the probability that a customer's purchase will be free (100% off)?
2. b. What is the probability that a customer will get no savings from the wheel (0% off)?
3. c. What is the probability that a customer will get at least 20% off?

To determine

To find: The probability that a customer’s purchase will be free (100% off).

Answer to Problem 1P

The probability that a customer’s purchase will be free (100% off) is 0.02.

Explanation of Solution

Given info:

The information is based on the “savings wheel”. Customer can get to spin the wheel after it stops, a pointer indicates how much they will save.

The wheel have 50 sections. In that 10 produce 0% off, 20 produce 10% off, 10 produce 20% off, 5 produce 30% off, 3 produce 40% off, 1 produce 50% off and 1 produce 100% off.

Calculation:

The formula for probability is,

$p(A)=\frac{\text{ Number of outcomes classified as }A}{\text{ Total number of possible outcomes}}$

The probability that a customer’s purchase will be free (100% off) is,

$\begin{array}{c}p(A)=\frac{\left(\begin{array}{l}\text{The number of sections that a customer's}\\ \text{ purchase will be free (100\% off )}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{1}{50} \\ =0.02\end{array}$

Thus, the probability value is 0.02.

To determine

To find: The probability that the number of sections that a customer’s will get no savings from the wheel (0 % off).

Answer to Problem 1P

The probability that the number of sections that a customer’s will get no savings from the wheel (0 % off) is 0.2.

Explanation of Solution

Calculation:

Calculating the probability that the number of sections that a customer’s will get no savings from the wheel (0 % off) is,

$\begin{array}{c}p(B)=\frac{\left(\begin{array}{l}\text{The number of sections that a customer's will }\\ \text{get no savings from the wheel (0\% off)}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{10}{50}\\ =\frac{1}{5}\\ =0.2\end{array}$

Thus, the probability is 0.2.

To determine

To find: The probability that a customer will get at least 20% off.

Answer to Problem 1P

The probability that a customer will get at least 20% off is 0.40.

Explanation of Solution

Calculation:

The probability that a customer will get at least 20% off is,

$\begin{array}{l}p\left(X\ge 20\%\right)=p\left(X=20\%\right)+p\left(X=30\%\right)+p\left(X=40\%\right)\\ +p\left(X=50\%\right)+p\left(X=100\%\right)\end{array}$

Calculating the individual probability values:

For 20:

$\begin{array}{l}p(X=20\%)=\frac{\left(\begin{array}{l}\text{The number of sections that a customer's }\\ \text{purchase will be 20\% off}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{10}{50}\end{array}$

For 30:

$\begin{array}{c}p\left(X=30\%\right)=\frac{\left(\begin{array}{l}\text{The number of sections that }\\ \text{ a customer's purchase will be 30\% off}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{5}{50}\end{array}$

For 40:

$\begin{array}{c}p\left(X=40\%\right)=\frac{\left(\begin{array}{l}\text{The number of sections that a customer's}\\ \text{ purchase will be 40\% off}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{3}{50}\end{array}$

For 50:

$\begin{array}{c}p\left(X=50\%\right)=\frac{\left(\begin{array}{l}\text{The number of sections that}\\ \text{a customer's purchase will be 50\% off}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{1}{50}\end{array}$

For 100:

$\begin{array}{c}p\left(X=100\%\right)=\frac{\left(\begin{array}{l}\text{The number of sections that }\\ \text{a customer's purchase will be 100\% off}\end{array}\right)}{\text{ Total number of sections }}\\ =\frac{1}{50}\end{array}$

Therefore, the required probability is,

$\begin{array}{c}p\left(X\ge 20\%\right)=\frac{\text{10}}{\text{50}}+\frac{\text{5}}{\text{50}}+\frac{\text{3}}{\text{50}}+\frac{\text{1}}{\text{50}}+\frac{\text{1}}{\text{50}}\\ =\frac{\text{20}}{\text{50}}\\ =0.40\end{array}$

Thus, the probability value is 0.40.