In the game Roulette, a ball spins on a circular wheel that is divided into 38 arcs of equal lenght, numbered 00, 0, 1, 2, ... , 35, 36. The number on the arc on which the ball stops is the outcome of one play of the game. The numbers are also colored as follows:

1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36 are red,
2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35 are black,
0, 00 are green
Define the following events:
A:{ Outcome is an even number (0 and 00 are considered neither odd nor even) }
B:{ Outcome is a red number }
C:{ Outcome is a green number }
D:{ Outcome is a low number (1-18) }
Find the following probablilities (Keep at least 3 digits) :
(a) P(A) =
0.473

(c) P(A U B U C)= 18+18+2/38= 1
I'm getting c wrong

P(A) is right.

P(C) = 2/38 = .053

Only two choices are green. Why did you add 18 twice?

it says 0.053 is wrong

To find the probability of event (c), P(A U B U C), we need to consider all the outcomes that satisfy at least one of the conditions: A, B, or C.

There are 18 red numbers, and since 0 and 00 are green, we have a total of 20 outcomes that satisfy condition B.

Condition A states that the outcome is an even number, so we need to count the total number of even numbers on the wheel. There are 18 black numbers and 2 green numbers (0 and 00), for a total of 20 outcomes that satisfy condition A.

Lastly, we need to consider the outcomes that satisfy condition C, which is the outcome being a green number. Since there are only 2 green outcomes (0 and 00), this condition is satisfied only by 2 outcomes.

To find the probability of P(A U B U C), we add up the number of outcomes that satisfy each condition and divide it by the total number of possible outcomes (38 in this case).

P(A U B U C) = (Number of outcomes that satisfy A or B or C) / (Total number of outcomes)
P(A U B U C) = (20 + 20 + 2) / 38
P(A U B U C) = 42 / 38
P(A U B U C) = 1.105

Therefore, the correct probability for P(A U B U C) is approximately 1.105

To find the probability of the union of events A, B, and C (A U B U C), we need to add the probabilities of each individual event and subtract the overlap (intersection) between them, since we don't want to count it twice.

Event A: Outcome is an even number (0 and 00 are neither odd nor even).
There are 18 even numbers out of the 38 possible outcomes, so P(A) = 18/38.

Event B: Outcome is a red number.
There are 18 red numbers out of the 38 possible outcomes, so P(B) = 18/38.

Event C: Outcome is a green number.
There are 2 green numbers (0 and 00) out of the 38 possible outcomes, so P(C) = 2/38.

To find the probability of the union of events A, B, and C (A U B U C), we can sum the probabilities of each event and subtract the intersection:
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

However, since A, B, and C are mutually exclusive (no overlap between them), the intersection probabilities are all 0:
P(A) = 18/38
P(B) = 18/38
P(C) = 2/38
P(A ∩ B) = 0
P(A ∩ C) = 0
P(B ∩ C) = 0
P(A ∩ B ∩ C) = 0

Therefore, the probability of the union of events A, B, and C is:
P(A U B U C) = P(A) + P(B) + P(C) = 18/38 + 18/38 + 2/38 = 38/38 = 1

So the correct value for P(A U B U C) is indeed 1, meaning that it is a certain event.