Question

Roulette An American roulette wheel has 38 slots with numbers $1$through$36,0,$and $00$, as shown in the figure. Of the numbered slots, $18$are red, $18$are black, and $2$—the $0$and $00$—are green. When the wheel is spun, a metal ball is dropped onto the middle of the wheel. If the wheel is balanced, the ball is equally likely to settle in any of the numbered slots. Imagine spinning a fair wheel once. Define events $B$: ball lands in a black slot, and $E$: ball lands in an even-numbered slot. (Treat$0$and $00$as even numbers.)

a. Make a two-way table that displays the sample space in terms of events $B$and $E$.

b. Find $P\left(B\right)$and$P\left(E\right)$.

c. Describe the event “$B$and $E$” in words. Then find the probability of this event.

d. Explain why $P(BorE)\ne P\left(B\right)+P\left(E\right)$. Then use the general addition rule to compute $P(BorE)$.

Short answer

a. A two way table displaying the sample space of events $B$and $E$.

Sample space of event $B$= $2,35,4,33,6,31,8,29,10,13,24,15,22,17,20,11,26,28$

Sample space of event $E$=$0,2,4,16,6,18,8,12,10,00,36,24,34,22,32,20,30,26,28$

b. The probability of event $B$= $0.4737$and event $E$=$0.5$

c. The probability of the event “$B$and $E$” = $0.2632$

d. The probability of the event “$B$or $E$” =$0.7105$

Step-by-step solution

Part (a) Step 1 : Given Information

We have to determine a two way table displaying the sample space of events $B$and $E$.

Part (a) Step 2 : Simplification

A roulette wheel has a total of $38$slots.

There are $18$red slots in all.

There are $18$black slots in all.

The number of green slots is two.

The event $B$is when a ball lands in a black slot.

The event $E$is when a ball lands in an even-numbered slot.

Sample space of event $B$= $2,35,4,33,6,31,8,29,10,13,24,15,22,17,20,11,26,28$

Sample space of event $E$= $0,2,4,16,6,18,8,12,10,00,36,24,34,22,32,20,30,26,28$

Part (b) Step 1 : Given Information

We have to determine the probability of event $B$and event $E$.

Part (b) Step 2 : Simplification

We fill use the following formula :-

$$\begin{gathered} P\left(B\right)=\frac{TotalnumberofoutcomesinthesamplespaceintheeventB}{Totalnumberofoutcomes} \\ P\left(E\right)=\frac{TotalnumberofoutcomesinthesamplespaceintheeventE}{Totalnumberofoutcomes} \end{gathered}$$

$$\begin{gathered} P\left(B\right)=\frac{18}{38}=0.4737 \\ P\left(E\right)=\frac{19}{38}=0.5 \end{gathered}$$

Part (c) Step 1 : Given Information

We have to determine the probability of the event “$B$ and $E$”

Part (c) Step 2 : Simplification

We will use the following formula :-

$P(B\cap E)=\frac{Totalnumberofoutcomesinthesamplespaceintheevent(B\cap E)}{Totalnumberofoutcomes}$

The ball lands in the black even-numbered slot in the event "$B$and $E$."

Even-numbered slots that are black = $2,4,6,8,10,24,22,20,26,28$

 $\hspace{1em}P(B\cap E)=\frac{10}{38}=0.2632$

Part (d) Step 1 : Given Information

We have to determine$P(BorE)$

Part (d) Step 2 : Simplification

The chance of a ball falling on the black slot or an even numbered slot is denoted by the letter $P(BorE)$.

Because$P(BandE)$is greater than zero, event $B$and event $E$are not mutually exclusive.

When one adds $P\left(B\right)$and $P\left(E\right)$, the$P(BandE)$are included twice.

As a result, $P(BorE)\ne P\left(B\right)+P\left(E\right)$

$$\begin{gathered} Fromtheabovesubparts,P\left(B\right)=0.4737,P\left(E\right)=0.5,P(BandE)=0.2632. \\ \hspace{1em}\hspace{1em}P(BorE)=0.4737+0.5–0.2632 \\ \hspace{1em}\hspace{1em}P(BorE)=0.7105 \end{gathered}$$