An American roulette wheel has $38$slots with numbers $1$through$36,0and00$as shown in the figure. Of the numbered slots, $18$are red, $18$are black, and $2-$the $0and00$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(B) and P(E).

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

d. Explain why P(B or E) ≠ P(B) + P(E). Then use the general addition rule to compute P(B or E).

Total number of slots in a roulette $=38$

Number of red slots $=18$

Number of black slots $=18$

Number of green slots$=2$

B is the event of a ball landing in a black slot

E is the event of a ball landing 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$E=0,2,4,16,6,18,8,10,00,36,24,34,22,32,20,30,26,28$

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

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

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

Event “B and E” means the ball lands in the black even numbered slot.

Slots which are black even numbered $=2,4,6,8,10,24,26,28$

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

$P(BorE)=P\left(B\right)+P\left(E\right)-P(BandE)$

P (B or E) means the probability of a ball landing on the black slot or on an even numbered.

As P (B and E) is not equal to zero, this means that event B and event E is not mutually exclusive.

By adding P (B) and P (E), one is including the P (B and E) twice.

Hence, P (B or E) P (B) + P (E) $P(BorE)\ne P\left(B\right)+P\left(E\right)$.

From the above sub parts, $P\left(B\right)=0.4737,P\left(E\right)=0.5,P(BandE)=0.6232$

$$\begin{gathered} P(BorE)=0.4737+0.5+0.2632 \\ P(BorE)=0.7105 \end{gathered}$$