A carnival wheel is split up into three colors: 62% blue, 20% red, and 18% green. Suppose we spin the wheel, observe the color that the pointer stops on, and r

Question

In a carnival game, a player spins a wheel that stops with the pointer on one (and only one) of three colors. The likelihood of the pointer landing on each color is as follows: 62 percent BLUE, 20 percent RED, and 18 percent GREEN. 

 

Suppose we spin the wheel, observe the color that the pointer stops on, and repeat the process until the pointer stops on RED. What is the probability that we will spin the wheel at least three times?

Answer 1

Hi Trent,

 

Think of the complementary event of NOT spinning the wheel at least 3 times. This would only happen if you spun at most 2 times. If you spun at most 2 times, then you either got red on the first spin or you got non-red (either blue or green) on the first spin followed by red on the second spin.

 

P (red on 1st) = 0.2

P (blue or green on 1st, followed by red on 2nd) = (0.62 + 0.18) × 0.2 = 0.16

 

This means

P (at most 2 spins) = 0.2 + 0.16 = 0.36

 

Finally,

P (at least 3 spins) = 1 - P (at most 2 spins) = 1 - 0.36 = 0.64