(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?

The correct answers are as follows.

a. 8.624%

b. 60.84%

c. 27.75%

Step-by-step explanation:

In order to find each of these, take the likelihood of each of steps and multiply them together.

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%

don't report but were is the digram

step-by-step explanation:

answer: #1 ads up by 8 and i don't know #2