Answer to Question #217966 in Statistics and Probability for Kojo

Question

  5.    Suppose that  

i. For what values of  are  and  mutually exclusive? For this value of , are

                          A and B independent?

           ii. For what values of , are  and  independent?

           Determine whether for this value of x, A and B are mutually exclusive.

 

Answer 1

$$8P(A\cup B)=5=>P(A\cup B)=\dfrac{5}{8}$$

$$\dfrac{2x}{P(A)}=1, P(B)=x$$

i. If $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0.$ Hence

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=P(A)+P(B)-0=P(A)+P(B)$$

$$\dfrac{5}{8}=2x+x$$

$$x=\dfrac{5}{24}$$

If $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B).$

Since $P(A\cap B)=0, P(A)\not=0, P(B)\not=0,$ then $A$ and $B$ are not independent.

ii.

If $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$

Then

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=P(A)+P(B)-P(A)P(B)$$

$$\dfrac{5}{8}=2x+x-x(2x)$$

$$16x^2-24x+5=0$$

$$x=\dfrac{24\pm\sqrt{(24)^2-4(16)(5)}}{2(16)}=\dfrac{3\pm2}{4}$$

Since $0\leq x\leq\dfrac{1}{2},$ we take $x=\dfrac{3-2}{4}=\dfrac{1}{4}.$

$$P(A)=\dfrac{1}{2}, P(B)=\dfrac{1}{4}, P(A\cap B)=\dfrac{1}{2}(\dfrac{1}{4})=\dfrac{1}{8}\not=0$$

Since $P(A\cap B)\not=0,$ then $A$ and $B$ are not mutually exclusive.