Q.3.23 - A First Course in Probability

Question

 Let A and B be events having positive probability. State whether each of the following statements is (I) necessarily true, (ii) necessarily false, or (iii) possibly true.

(a) If A and B are mutually exclusive, then they are independent.

(b) If A and B are independent, then they are mutually exclusive.

(c)$P\left(A\right)=P\left(B\right)=.6$, and A and B are mutually exclusive.

(d) $P\left(A\right)=P\left(B\right)=.6,$ and A and B are independent.

Step-by-Step Solution

Answer

a)A and B are mutually exclusive, then they are independent is Necessarily false

b) Necessarily false - use the characterization of independence with probability of intersection

c)  calculate $P(A\cup B)$Necessarily false

d)  $P\left(A\right)=P\left(B\right)=.6$ A and B are independent is Possible

Step1: A and B are mutually exclusive, then they are independent (part a)

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

a)$AB=\varnothing \Rightarrow A$ and B are independent

Necessarily false

Because

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

In this case

$AB=\varnothing \Rightarrow P\left(AB\right)=0\ne \stackrel{>0}{\overbrace{P\left(A\right)P\left(B\right)}}$

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

Step2: A and B are independent, then they are mutually exclusive (part b)

b) If A and B are independent $\Rightarrow AB=\varnothing$

Necessarily false

Because

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

In this case

A and B are independent $\Rightarrow P\left(AB\right)=P\left(A\right)P\left(B\right)>0\Rightarrow AB\ne \varnothing$

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

Step3:   A B = ∅  and  P ( A ) = P ( B ) = 0.6 (part c)

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

Necessarily false

If $A\cap B=\varnothing$ then:

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

This fact is stated in the third probability axiom.

Incorporating the second assumption would imply:

$P(A\cup B)=0.6+0.6=1.2$

And it is well known$A\cup B$ that is an event, and the probability of any event should be considered$<1$

That's why there's a contradiction here.

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

Step4: Find A and B independent (part d)

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 $1,2 or 3$.

B - the second number is$1,2 or 3 .$

Then

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

and 

A and B are separate entities.