In a carnival game, a player spins a wheel that stops with the pointer on one (a
ID: 3179398 • Letter: I
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: 61 percent BLUE, 20 percent RED, and 19 percent GREEN.
(a) Suppose we spin the wheel, observe the color that the pointer stops on, and repeat the process until the pointer stops on BLUE. What is the probability that we will spin the wheel exactly three times?
(b) 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?
(c) Suppose we spin the wheel, observe the color that the pointer stops on, and repeat the process until the pointer stops on GREEN. What is the probability that we will spin the wheel 2 or fewer times?
Explanation / Answer
a) P(Blue) = 0.61
So,
P(Not Blue) = 1 - 0.61 = 0.39
P(Wheel spin exactly three times)
= P(Not blue for the first two spins)*P(Blue on the third spin)
= (0.39)2*(0.61)
= 0.092781
b) P(Red) = 0.2
P(Not Red) = 1 - 0.2 = 0.8
P(Wheel spin atleast three times)
= 1 - P(Wheel spin less than 3 times)
= 1 - [P(1 time spin) + P(2 time spin)]
= 1 - [0.2 + (0.8)*(0.2)]
= 1 - 0.36
= 0.64
c) P(Green) = 0.19
P(Not Green) = 1 - 0.19 = 0.81
P(Wheel spin 2 or fewer times)
= P(1 time spin) + P(2 time spin)
= 0.19 + (0.81)*(0.19)
= 0.3439