91Ó°ÊÓ

$P\left(A\right)>0,P\left(B\right)>0$

a)$AB=âˆ\dots ⇒A$and B are independent

Necessarily false

Because

A and B are independent $⇔P\left(AB\right)=P\left(A\right)P\left(B\right)$

In this case

$AB=\mathrm{âˆ\dots }⇒P\left(AB\right)=0≠\stackrel{>0}{\stackrel{Áåž}{P\left(A\right)P\left(B\right)}}$

As a result, A and B are not necessarily independent.

b) If A and B are independent $⇒AB=âˆ\dots$

Necessarily false

Because

A and B are independent $⇔P\left(AB\right)=P\left(A\right)P\left(B\right)$

In this case

A and B are independent $⇒P\left(AB\right)=P\left(A\right)P\left(B\right)>0⇒AB≠âˆ\dots$

As a result, A and B have a non-empty intersection.

$\text{c)}AB=\mathrm{âˆ\dots }\text{and}P\left(A\right)=P\left(B\right)=0.6$

Necessarily false

If $Aâˆ\copyright B=âˆ\dots$then:

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

This fact is stated in the third probability axiom.

Incorporating the second assumption would imply:

$P(A∪B)=0.6+0.6=1.2$

And it is well knownlocalid="1649651381942" $A∪B$that is an event, and the probability of any event should be consideredlocalid="1649651386336" $<1$

That's why there's a contradiction here.

d) A and B independent and localid="1649651391474" $P\left(A\right)=P\left(B\right)=0.6$

Possible - this is demonstrated by an example in which the statement is true.

Choose two numbers at random in$\{1,2,3,4,5\}$, independently.

A - the first number is $1,2or3$.

B - the second number is$1,2or3.$

Then

$P\left(A\right)=P\left(B\right)=0.6$

and

A and B are separate entities.