Question

Roulette is a very popular game in many American casinos. In roulette, a ball spins on a circular wheel that is divided into 38 arcs of equal length, bearing the numbers 00, 0, 1, 2, ..., 35, 36. The number of the arc on which the ball stops is the outcome of one play of the game. The numbers are also colored in the following manner:

Red: 1,3,5,7,9,12,14,16,18,19,21.23,25,27,30,32.34,36
Black: 2.4,6,8,10,11,13,15,17,20,22,24.26,28,29,31,33,35 Greerz: 00,O
Players may place bets on the table in a variety of ways, including bets on odd, even, red, black, high, low, etc. Define the following events:
A: (Outcome is an odd number (00 and 0 are considered neither odd nor even)]
B: (Outcome is a black number]
C: {Outcome is a low number (1-18))
Calculate the probabilities of the following events:
a. AUB
b. A ∩ C
c. BU C
d. Bc
e. A ∩ B ∩ C

Answer 1

a. P(AUB) = 0.74

b. P(A∩C) = 0.24

c. P(BUC) = 0.71

d. P(Bc) = 0.55

e. P(A∩B∩C) = 0.11

Explanation:

n(U) = Total number of elements in the set = number of elements in the universal set = 38

A: (Outcome is an odd number (00 and 0 are considered neither odd nor even)]

An odd number referred to any integer, that is not a fraction, which is not possible to be divided exactly by 2. Therefore, we have:

A: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35

B: 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35

C: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

a. Calculate the probability of AUB

AUB picks not more than one each of the elements in both A and B without repetition. Therefore, we have:

AUB = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 33, 35

n(AUB) = number of elements in AUB = 28

P(AUB) = probability of AUB = n(AUB) / n(U) = 28 / 38 = 0.736842105263158 = 0.74

b. Calculate the probability of A ∩ C

A∩C picks only the elements that are common to both A and C. Therefore, we have:

A∩C: 1, 3, 5, 7, 9, 11, 13, 15, 17

n(A∩C) = number of elements in A∩C = 9

P(A∩C) = probability of A∩C = n(A∩C) / n(U) = 9 / 38 = 0.236842105263158 = 0.24

c. Calculate the probability of BUC

BUC picks not more than one each of the elements in both B and C without repetition. Therefore, we have:

BUC: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35

n(BUC) = number of elements in BUC = 27

P(BUC) = probability of A∩C = n(BUC) / n(U) = 27 / 38 = 0.710526315789474 = 0.71

d. Calculate the probability of Bc

Bc indicates B component it represents all the elements in the universal set excluding the elements in B. Therefore, we have:

Bc: 00, 0, 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36

n(Bc) = number of elements in Bc = 20

P(Bc) = probability of Bc = n(Bc) / n(U) = 20 / 38 = 0.526315789473684 = 0.55

e. Calculate the probability of A∩B∩C

A∩B∩C picks only the elements that are common to A, B and C. Therefore, we have:

A∩B∩C: 11, 13, 15, 17

n(Bc) = number of elements in A∩B∩C = 4

P(A∩B∩C) = probability of A∩B∩C = n(A∩B∩C) / n(U) = 4 / 38 = 0.105263157894737 = 0.11

Note: All the answers are rounded to 2 decimal places.