In a version of the game of roulette, a steel ball is rolled onto a wheel that contains 19 red, 19 black and 2 green slots. if the ball is rolled 25 times, the probability that it falls into the green slots two or more times:

The probability is 0.3576

Step-by-step explanation:

The probability for the ball to fall into the green ball in one roll is 2/1919+2 = 2/40 = 1/20. The probability for the ball to roll into other color is, therefore, 19/20.

For 25 rolls, the probability for the ball to never fall into the green color is obteined by powering 19/20 25 times, hence it is 19/20^25 = 0.2773

To obtain the probability of the ball to fall once into the green color, we need to multiply 1/20 by 19/20 powered 24 times, and then multiply by 25 (this corresponds on the total possible positions for the green roll). The result is 1/20* (19/20)^24 *25 = 0.3649

The exercise is asking us the probability for the ball to fall into the green color at least twice. We can calculate it by substracting from 1 the probability of the complementary event: the event in which the ball falls only once or 0 times. That probability is obtained from summing the disjoint events: the probability for the ball falling once and the probability of the ball never falling. We alredy computed those probabilities.

As a result. The probability that the ball falls into the green slot at least twice is 1- 0.2773-0.3629 = 0.3576

me! what did you put for this question?

step-by-step explanation: