91Ó°ÊÓ

The formula is used $\mathbf{P}\left(A|B\right)\mathbf{=}\frac{\mathbf{P}\left(Aâˆ\copyright B\right)}{\mathbf{P}\left(B\right)}$.

$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{∪}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$

Here,

$P\left(1\right)=.3,P\left(2\right)=.2,P\left(3\right)=.1,P\left(4\right)=.1,P\left(5\right)=.1$, and $P\left(6\right)=.2$.

$$\begin{gathered} P(Aâˆ\copyright B)=P\left(0\right) \\ =0 \end{gathered}$$

role="math" localid="1660215728600" $$\begin{gathered} P(Bâˆ\copyright C)=P\left(2\right) \\ =0.2 \end{gathered}$$

$$\begin{gathered} P(Aâˆ\copyright C)=P\left(1\right)+\left(2\right)+P\left(3\right)+P\left(5\right)+P\left(6\right) \\ =0.3+0.2+0.1+0.1+0.2 \\ =0.9 \end{gathered}$$

$$\begin{gathered} P(A∪B∪C)=P\left(1\right)+\left(2\right)+P\left(3\right)+P\left(5\right)+P\left(6\right) \\ =0.3+0.2+0.1+0.1+0.2 \\ =1 \end{gathered}$$

$$\begin{gathered} P\left(B^c\right)=1-P\left(B\right) \\ =1-\left(P\left(2\right)+P\left(4\right)\right) \\ =1-\left(0.2+0.1\right) \\ =0.7 \end{gathered}$$

$$\begin{gathered} P(A^câˆ\copyright B)=P\left(2\right)+P\left(4\right) \\ =0.2+0.1 \\ =0.3 \end{gathered}$$

$$\begin{gathered} P(B\backslash C)=\frac{P\left(2\right)}{P\left(2\right)+P\left(5\right)+P\left(6\right)} \\ =\frac{0.2}{0.2+0.1+0.2} \\ =0.4 \end{gathered}$$

$$\begin{gathered} P\left(B\backslash A\right)=\frac{P\left(0\right)}{P\left(1\right)+P\left(3\right)+P\left(5\right)} \\ =\frac{0}{0.3+0.1+0.1} \\ =0 \end{gathered}$$

Hence, the values are 0, 0.2, 0.9, 1, 0.7, 0.3, 0.4, 0 respectively.

b.

No, the both events are not independent, and they are mutually exclusive, because$P\left(Aâˆ\copyright B\right)=0$

So, Events are independent but not mutually exclusive.

c.

No, the both events are not independent, mutually exclusive, because

$P(Bâˆ\copyright C)=0.2$and$P(Bâˆ\copyright C)≠P\left(B\right)â‹\dots P\left(C\right)$

Therefore, both events are not independent and mutually exclusive.