(20 points) 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, 21 percent RED, and 18 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?

Part A =  9.278%

Part B = 62.41%

Part C = 32.76%

Step-by-step explanation:

Let  

B = the chance for get blue= 61% = 0.61,  

R = the chance for get red            =0.21  

G= the chance for get green        = 0.18

Part A

We want to know the chance of getting blue exactly at the third spins. That means we have to get non-blue (red or green) for the first 2 spins then get blue at third. The negation of B will be equal to 1-0.61= 0.39. The calculation will be:  

P(B') * P(B') * P(B) = (1-0.61) * (1-0.61) * 0.61= 0.092781= 9.278%

Part B

We want to know the chance of getting red at least after three spins. Take notes that part B use "at least" while part A use "exactly". The number of possible ways that met the condition will be infinite since we can roll forever if red never comes. We should find the probability of the negation instead. The condition that doesn't fulfill at least three spins is that 2 spins or 1 spin. Two spins mean you get non-red at first, then get red afterward.

The calculation will be:

P(R') * P(R) = (1-0.21) * 0.21= 0.1659

One spin means you get red at first spin, the chance will be P(R)= 0.21.

The negation of both condition will be:

P(>3)= ~(P(2) + P(1))= 1 - (0.1659+0.21) = 0.6241= 62.41%

Part C

We want to know the chance of getting green in 2 or fewer spins. It similar to part B, we need to know the probability of getting green at 2nd spin and 1st spin.

The chance for getting green at 2nd spin will be:

P(G') * P(G) = (1-0.18) * 0.1821= 0.1476

The chance for getting green at 1st spin will be P(G)= 0.18

The chance for 2 or fewer spins will be:

P(<2) = P(1) + P(2) = 0.18+ 0.1476 =0.3276 = 32.76%