91影视

We have to determine 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 $E$= $0,2,4,16,6,18,8,12,10,00,36,24,34,22,32,20,30,26,28$

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

We fill use the following formula :-

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

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

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

We will use the following formula :-

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

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

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

鈥�$鈥�P(B鈭�E)=\frac{10}{38}=0.2632$

We have to determine$P(BorE)$

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

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