Question

American roulette is a game in which a wheel turns on a spindle and is divided into 38 pockets. Thirty-six of the pockets are numbered 1−36, of which half are red and half are black. Two of the pockets are green and are numbered 0 and 00 (see figure). The dealer spins the wheel and a small ball in opposite directions. As the ball slows to a stop, it has an equal probability of landing in any of the numbered pockets.

(a) Find the probability of landing in the number 0 pocket.


(b) Find the probability of landing in a black pocket.


(c) Find the probability of landing in a green pocket or a red pocket.


(d) Find the probability of landing in the number 6 pocket on two consecutive spins.


(e) Find the probability of landing in a red pocket on three consecutive spins.

Answer 1

(a) Since there are 38 pockets and only one of them is 0, the probability of landing in the number 0 pocket is 1/38.

(b) Since half of the numbered pockets are black, and there are 36 numbered pockets, the probability of landing in a black pocket is 18/38 or 9/19.

(c) The probability of landing in a green pocket is 2/38 or 1/19, and the probability of landing in a red pocket is 18/38 or 9/19. To find the probability of landing in a green pocket or a red pocket, we add these probabilities: 1/19 + 9/19 = 10/19.

(d) The probability of landing in the number 6 pocket on one spin is 1/38. Since we want this to happen on two consecutive spins, we multiply this probability by itself: (1/38) × (1/38) = 1/1444.

(e) The probability of landing in a red pocket on one spin is 9/19. Since we want this to happen on three consecutive spins, we multiply this probability by itself three times: (9/19) × (9/19) × (9/19) = 729/6859 or approximately 0.1063.