**7.3 Probability of Mutually Exclusive Events**

**1.**Two events are

**if they**

*mutually exclusive***cannot occur**at the

**same time**.

**2.**If

*A*and

*B*are mutually exclusive events, then

P(A υ B) = P(A) + P(B) |

**Example:**

A
bag contains 3 blue cards, 4 green cards and 5 yellow cards. A card is chosen
at random from the box. Find the probability that the chosen card is green or
yellow.

*Solution:*
Let

*G*= event when a green card is chosen.*Y*= event when a yellow card is chosen.

The
sample space,

*S*= 12,*n*(*S*) = 12*n*(

*G*) = 4 and

*n*(

*Y*) = 5

$\begin{array}{l}P(G)=\frac{n\left(G\right)}{n\left(S\right)}=\frac{4}{12}\\ P(Y)=\frac{n\left(Y\right)}{n\left(S\right)}=\frac{5}{12}\end{array}$

Events

$\begin{array}{l}G\cap Y=\varnothing \\ P(G\cup Y)=\text{}P(G)+P(Y)\\ \text{}=\frac{4}{12}+\frac{5}{12}\\ \text{}=\frac{9}{12}=\frac{3}{4}\end{array}$

*G*and*Y*cannot occur simultaneously because we cannot obtain green card and yellow card at the same time. Therefore, events*G*and*Y*are**mutually exclusive**.$\begin{array}{l}G\cap Y=\varnothing \\ P(G\cup Y)=\text{}P(G)+P(Y)\\ \text{}=\frac{4}{12}+\frac{5}{12}\\ \text{}=\frac{9}{12}=\frac{3}{4}\end{array}$