Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of $38$numbers, and each number is assigned to a color and a range.

a. List the sample space of the $38$possible outcomes in roulette.

b. You bet on red. Find P(red).

c. You bet on -$1^{st}12$- (1st Dozen). Find $P\left(1^{st}12\right)$.

d. You bet on an even number. Find P(even number).

e. Is getting an odd number the complement of getting an even number? Why?

f. Find two mutually exclusive events.

g. Are the events Even and $1^{st}$Dozen independent?

There are $18$red numbers and $18$odd numbers.

The sample space of the $38$possible outcomes in the given game is,

Let it be s.

$S=\left\{\begin{array}{l}0,00,1,2,3,4,5,6,7,8\\ 9,10,11,12,13,14,15\\ 16,17,18,19,20,21,22\\ 23,24,25,26,27,28,29\\ 30,31,32,33,34,35,36\end{array}\right\}$

There are $18$red numbers and $18$odd numbers.

The probability of $P\left(\text{red}\right)$is

$$\begin{gathered} P\left(red\right)=\frac{n\left(red\right)}{n\left(s\right)} \\ n\left(\text{red}\right)=18;n\left(S\right)=38 \\ P\left(red\right)=\frac{18}{38} \\ P\left(red\right)=0.47 \end{gathered}$$

There are $18$red numbers and $18$odd numbers.

The probability of $P\left(1^{st}12\right)$is

$$\begin{gathered} P\left(1^{st}12\right)=\frac{n\left(1^{st}12\right)}{n\left(S\right)} \\ n\left(1^{st}12\right)=12;n\left(S\right)=38 \\ P\left(1^{st}12\right)=\frac{12}{38} \\ P\left(1^{st}12\right)=0.32 \end{gathered}$$

There are $18$ red numbers and $18$ odd numbers.

The probability of $P\left(evennumber\right)$is

$$\begin{gathered} P\left(\text{even number}\right)=\frac{n\left(\text{evennumber}\right)}{n\left(S\right)} \\ n\left(\text{evennumber}\right)=18;n\left(S\right)=38 \\ P\left(\text{even number}\right)=\frac{18}{38} \\ P\left(\text{even number}\right)=0.47 \end{gathered}$$

There are $18$ red numbers and $18$ odd numbers.

Getting an odd number is not the complement of getting an even number because we have $38$outcomes in the sample space which includes $0$and $00$.

There are $18$ red numbers and $18$ odd numbers.

To get the mutually exclusive events, there should not be any common terms in between the two events.

Therefore, the two mutually exclusive events are odd and even numbers or in other words, it is red and black numbers.

There are $18$ red numbers and $18$ odd numbers.

To get two independent even, we have to show the following.

$$\begin{gathered} P(\text{Even}\mid \text{1stdozen})=P\left(\text{Even}\right) \\ P(\text{lstdozen}\mid \text{Even})=P\left(1\text{stdozen}\right) \\ P\left(\text{Even AND 1stdozen}\right)=P\left(\text{Even}\right)P\left(1\text{stdozen}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} P(\text{Even}\mid 1\text{stdozen})=0.5;P\left(\text{Even}\right)=0.47 \\ P(\text{Even}\mid 1\text{stdozen})\ne P\left(\text{Even}\right) \end{gathered}$$