(4 pts) 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: 63 percent BLUE, 22 percent RED, and 15 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?

a. For this one, we need to note that the first two spins should not be blue. The likelihood of them not being blue is the chance for red and green put together (37%). Then we want the third spin to be the blue odds (63%)

.37 * .37 * .63 = .08624 = 8.624%

b. For the purpose of this, we are looking just for the first two spins to not be red (78% probability). After that, it does not matter what the third spin is since it says "at least 3 times)

.78 * .78 = .6084 = 60.84%

c. For this one, we need to combine two things. Firstly, the chance that the first spin is green (15%). Then we would add in the likelihood that the first spin is not green (85%), but the second is.

.15 + (.85 * .15) = .2775 = 27.75%

fichoh fichoh · Answer 2 · 2021-10-04T08:20:08+02:00

The probabilities of obtaining the required event based on the likelihood of the pointer stopping on each color are 0.086247, 0.6084 and 0.2775 respectively

Probability of Blue ; P(B) = 0.63

Probability of Red ; P(R) = 0.22

Probability of Green ; P(G) = 0.15

Probability of Blue after exactly 3 spins :

1st spin = P(not B) = P(R)+P(G) = (0.22+0.15) = 0.37

2nd spin = P(not B) = P(R)+P(G) = (0.22+0.15) = 0.37

3rd spin = P(B) = 0.63

P(R exactly after 3 spins) = (0.37 × 0.37 × 0.63) = 0.086247

2.)

Probability of spinning atleast 3 times before obtaining RED;

P(not R) on first two spins ;

P(not R) = P(B) + P(G) (0.63+0.15) = 0.78

P(not R) × P(not R) = 0.78 × 0.78 = 0.6084

3.)

Probability of obtaining Green after 2 or fewer spins :

P(Green after 2 spins) + P(Green on first spin)

P(not G) = 0.63 + 0.22 = 0.85

[P(not G) × P(G)] + P(G)

[(0.85 × 0.15) + 0.15]

(0.1275 + 0.15)

= 0.2775

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