Question

Suppose that A and B are mutually exclusive events.

Part (a) Use the special addition rule to express P(A or B) in terms of P(A) and P(B)

Part (b) Show that the general addition rule gives the same answer as that in part (a)

Short answer

Part (a) $P(AorB)=P\left(A\right)+P\left(B\right)$

Part(b) As a result, the general addition rule yields the same result as component $A$'s special addition rule.

Step-by-step solution

Part (a) Step 1. Given information.  

Consider the occurrences $A$and$B$ as mutually exclusive.

Part(a) Step 2. If events A and B are mutually exclusive.

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

Part (b) Step 1. Given information.  

Consider the occurrences $A$ and $B$ as mutually exclusive.

Part (b) Step 2. From the given data

$P(AandB)=0$

role="math" localid="1651246672127" $$\begin{gathered} P(AorB)=P\left(A\right)+P\left(B\right)-P(AandB) \\ P\left(A\right)+P\left(B\right)-0 \\ P\left(A\right)+P\left(B\right) \end{gathered}$$

As a result, the general addition rule yields the same result as component $A$'s special addition rule.

Part (a) Step 1. Given information.  

Consider the occurrences A and B as mutually exclusive.

Part(a) Step 2. If events A and B are mutually exclusive.

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

Part (b) Step 1. Given information.  

Consider the occurrences A and B as mutually exclusive.

Part (b) Step 2. From the given data

$P(AandB)=0$

$$\begin{gathered} P(AorB)=P\left(A\right)+P\left(B\right)-P(AandB) \\ P\left(A\right)+P\left(B\right)-0 \\ P\left(A\right)+P\left(B\right) \end{gathered}$$

As a result, the general addition rule yields the same result as component$A$'s special addition rule.