Question

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,

a. What is the probability that a customer’s purchase will be free (100% off)?

b. What is the probability that a customer will get no savings from the wheel

(0% off)?

c. What is the probability that a customer will get at least 20% off?

Answer 1

Probability distribution of “Savings Wheel” is

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

P(Customer’s purchase will be 100% off) $= \frac{Number \; of sections \;will \; get \; 100 \; \% \; off}{Total \; number \; of \; sections \; in \; the \; wheel}$

$= \frac{1}{50} \\ = 0.02$

b. The probability of a customer’s purchase will get no savings from the wheel (0% off) is

P(Customer’s purchase will get no savings from the wheel (0% off)) $= \frac{Number \; of sections \;will \; get \; 0 \; \% \; off}{Total \; number \; of \; sections \; in \; the \; wheel}$

$= \frac{10}{50} \\ = 0.2$

(c) The probability of a customer will get at least 20% off is

P(Customer will get at least 20% off) $= \frac{10}{50}+ \frac{5}{50}+\frac{3}{50}+ \frac{1}{50} + \frac{1}{50}$

$= 0.2 + 0.1 + 0.06 + 0.02 + 0.02 \\ = 0.4$