Chapter 10.2, Problem 30E

(a)

To determine

To explain:

The reason behind the similarity probability of event $A$ and $B$ for the following:

  $P\left(B\right)=P\left(B|A\right)$ and $P\left(A\right)=P\left(A|B\right)$

Answer to Problem 30E

With the help of conditional and joint distributions the given independent events are justified.

Explanation of Solution

Given information:

Event $A$ and $B$ are independent events

Check for $P\left(B\right)=P\left(B|A\right)$

Take the right hand side of the above equation:

  $\begin{array}{l}P\left(B|A\right)=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ P\left(B|A\right)=\frac{P\left(B\right)\times P\left(A\right)}{P\left(A\right)}\\ P\left(B|A\right)=P\left(B\right)\end{array}$

Here, right hand side is equal to left hand side. Hence, proved that the event $A$ and $B$ are independent events.

Check for $P\left(A\right)=P\left(A|B\right)$

Take the right hand side of the above equation:

  $\begin{array}{l}P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ P\left(A|B\right)=\frac{P\left(A\right)\times P\left(B\right)}{P\left(B\right)}\\ P\left(A|B\right)=P\left(A\right)\end{array}$

Here, right hand side is equal to left hand side. Therefore the given equation is correct for the conditional probability.

(b)

To determine

To explain:

Probability of event $A$ and $B$ is same as $P\left(B\right)\cdot P\left(A|B\right)$ or not

Answer to Problem 30E

Yes, Probability of event $A$ and $B$ is same as $P\left(B\right)\cdot P\left(A|B\right)$ and it shown as $P\left(A\cap B\right)=P\left(B\right)\cdot P\left(A|B\right)$

Explanation of Solution

Given information:

Event $A$ and $B$ are independent events

For the given statement the conditional probability is considered as $P\left(A|B\right)$ .

Now, conditional probability expressed as:

  $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

Here, need to express the probability for both events $A$ and $B$ . Let’s rephrase the above equation as:

  $P\left(A\cap B\right)=P\left(B\right)\cdot P\left(A|B\right)$

Therefore, the given justification statement is correct.