Suppose a player observes 3 players of roulette. Note: A roulette wheel has 18 red numbers, 18 black numbers, and 2 green numbers. The results of different plays on a roulette wheel are independent of one another.
a) what is the chance the ball will land in a red slot all 3 plays?
b) what is the chance the ball will land in the same color slot all 3 plays?
c) what is the chance the 4th play will result in red number, given the first 3 plays were all black?

(a) There are 38 possible outcomes of one spin. 18 of them are red. Therefore, the probability of getting red on any ONE spin is 18/38 0.474. If you want to know the probability of sequential events happening, it's the product of the probabilities of each individual event. Hence P(three reds) = P(red)*P(red)*P(red) = (18/38)*(18/38)*(18/38) 0.106.

(b) There are three possible "all the same color" scenarios - all three on red, all three on black, or all three on green. The total probability of one of these scenarios OR another OR the other, is the sum of the probabilities of each. We already saw in part (a) that the probability of all reds is around 10.6%. What's the probability of all blacks? (That one's easy since there are the same number of blacks as reds). And how about the probability of all greens? Just add up the three probabilities and there's your answer.

(c) The roulette wheel doesn't know or care what colors have already come up before. It doesn't matter whether the last three spins were black, red, green, or any combination thereof. The probability of any spin coming up red is always 18/38.