91影视

Given that$P\left(C\right)=0.4,P\left(D\right)=0.5andP(C鈭�D)=0.6$.

$P\left(CandD\right)$is calculated as

$P\left(CandD\right)=P\left(C|D\right)P\left(D\right)$

Substituting the values, we get

$$\begin{gathered} P\left(CandD\right)=0.6脳0.5 \\ P\left(CandD\right)=0.30 \end{gathered}$$

Given that$P\left(C\right)=0.4,P\left(D\right)=0.5andP(C鈭�D)=0.6$.

We have

$$\begin{gathered} P\left(CandD\right)=0.30 \\ \text{Thus},P\left(CANDD\right)鈮�0 \end{gathered}$$

Therefore, events C and D are not mutually exclusive.

Given that$P\left(C\right)=0.4,P\left(D\right)=0.5andP(C鈭�D)=0.6$.

We have $P\left(C\right)=0.4$

$P\left(C|D\right)=0.6$

Thus,$P\left(C\right)鈮�P(C鈭�D)$

The first condition is not satisfied.

Therefore, events CandD are not independent.

Given that$P\left(C\right)=0.4,P\left(D\right)=0.5andP(C鈭�D)=0.6$

$P\left(CorD\right)$is calculated as

$P(C\text{OR}D)=P\left(C\right)+P\left(D\right)-P(C\text{AND}D)$

Substituting the values, we get

$$\begin{gathered} P(CORD)=0.4+0.5-0.3 \\ P(CORD)=0.9-0.3 \\ P(CORD)=0.6 \end{gathered}$$

Given that$P\left(C\right)=0.4,P\left(D\right)=0.5andP(C鈭�D)=0.6$.

$P\left(D|C\right)$is calculated as

$$\begin{gathered} P\left(DANDC\right)=P(D鈭�C)P\left(C\right) \\ P(D鈭�C)=\frac{P\left(DANDC\right)}{P\left(C\right)} \end{gathered}$$

Substituting the values, we get

$$\begin{gathered} P\left(D|C\right)=\frac{0.3}{0.4} \\ P\left(D|C\right)=0.75 \end{gathered}$$