# WINNING BIG!

Date of the Problem
August 08, 2022

On the game show Wheel of Fortune®, contestants spin a wheel, like the one shown here, to determine how much money will be won for solving the puzzle. What is the probability that a contestant spins the wheel and it stops on a space that either is not purple or is worth more than \$500? Express your answer as a common fraction.

The probability that, once spun, the wheel stops on a space that is not purple is 20/24 = 5/6. The probability that the wheel stops on a space that is worth more than \$500 is 10/24 = 5/12. Seven of the spaces that are worth more than \$500 are also not purple and we need to account for these duplicates. Thus, the probability of the wheel stopping on a space that either is not purple or a space worth more than \$500 is 5/6 + 5/12 – 7/24 = (20 + 10 – 7)/24 = 23/24.

In how many ways with multiple spins can a contestant spin a sum of \$1200 with the wheel stopping on spaces of a single color and stopping on a space no more than twice? (The order of the spins matters.)

First, we see that with each of the colors, it is possible to get a sum of \$1200 in multiple spins. However, to get a sum of \$1200 in a single color with blue or orange would require the wheel to stop on a space more than twice. So, let’s examine ways to get a sum of \$1200 with the wheel stopping on green, red, pink and purple spaces.

Green

\$600 × 2: Since there is 1 green \$600, this can happen in 1 way.

\$300 × 2 + \$600: Since there is 1 green \$300 and 1 green \$600, this can happen in 3 ways.

Red

\$300 + \$900: Since there is 1 red \$300 and 1 red \$900, this can happen in 2 ways.

Pink

\$300 + \$900: Since there is 1 pink \$300 and 1 pink \$900, this can happen in 2 ways.

\$450 × 2 + \$300: Since there is 1 pink \$450 and 1 pink \$300, this can happen in 3 ways.

Yellow

\$400 × 2 + \$400: Since there are 2 yellow \$400 sections, let’s call one A and one B. The possible spin options to get \$1200, then, are ABB, BAB, BBA, BAA, ABA, AAB. Thus, this can happen in 6 ways.

Purple

\$600 × 2: Since there are 2 purple spaces with \$600, this can happen in 4 ways.

That means a contestant can spin a sum of \$1200 with the wheel stopping on spaces of a single color and stopping on a space no more than twice in 1 + 3 + 2 + 2 + 3 + 6 + 4 = 21 ways.

What is the probability that a contestant spins the wheel three times and it stops on a different color each time with the same dollar amount? Express your answer as a common fraction.

The only amount that appears on the wheel in 3 or more different colors is \$300. There is 1 red, 1 pink, 1 green, 1 blue and 1 orange space labeled \$300. From these, we see that there are 5C3 = 10 combinations of 3 different colors, which can each occur in 3! different ways. That’s a total of 10 × 6 = 60 ways. There are 243 = 13,824 results when the wheel is spun three times. That means the probability of the wheel stopping on three different colors with the same dollar amount in three spins is 60/13,824 = 5/1152. You can also consider that on the first spin, 5/24 spaces will work. On the second spin, 4/24 spaces will work. And on the third spin, only 3/24 spaces work. The probability is 5/24 × 4/24 × 3/24 = 5/2 × 1/24 × 1/24 = 5/1152.

Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.

Page 2 contains ONLY PROBLEMS. ♦

Math topic
CCSS (Common Core State Standard)
Difficulty