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, 22 percent RED, and 17 percent GREEN. Note: Your answers should be rounded to three decimal places.

(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?

Answer #1

a)

Required probability = P(we will spin the wheel exactly three times) = P(the pointer stops for the first time at Blue on 3rd spin) = (1-0.61)(1-0.61)(0.61) = 0.093

b)

Required probability = P( we will spin the wheel at least three times) = 1 - P( we will spin the wheel less than three times)

= 1 - P( we will spin the wheel once) - P( we will spin the wheel twice)

= 1 - (0.22) - (1-0.22)(0.22) = 0.608

c)

Required probability = P( we will spin the wheel 2 or fewer times)

= P( we will spin the wheel once) + P( we will spin the wheel twice)

= 0.17 + (1-0.17)(0.17) = 0.311

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