**7.2 Probability of the Combination of Two Events**

**1.**For two events,

*A*and

*B*, in a sample space

*S*, the events

*A*∩

*B*(

*A*and

*B*) and

*A*υ

*B*

(

*A*or

*B*) are known as combined events.

**2.**The probability of the

**of sets**

*union**A*and

*B*is given by:

$\overline{)\text{}P(A\cup B)=\text{}P(A)+P(B)-P(A\text{}\cap \text{}B)\text{}}$

**3.**The probability of the union of sets

*A*and

*B*can also be calculated using an alternative method, i.e.

$\overline{)\text{}P(A\cup B)=\frac{n(A\cup B)}{n(S)}\text{}}$

**4.**The probability of event

*A*and event

*B*occurring,

*P*(

*A*∩

*B*) can be determined by the following formula.

$\overline{)\text{}P(A\cap B)=\frac{n(A\cap B)}{n(S)}\text{}}$

**Example:**

Given
a universal set ξ = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. A number
is chosen at random from the set ξ . Find the probability that

(a)
an even number is chosen.

(b)
an odd number or a prime number is chosen.

*Solution:*
The
sample space,

*S*= ξ*n*(

*S*) = 14

**(a)**

Let

*A*= Event of an even number is chosen*A*= {2, 4, 6, 8, 10, 12, 14}

*n*(

*A*) = 7

$\begin{array}{l}P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ \text{}=\frac{7}{14}=\frac{1}{2}\end{array}$

**(b)**Let,

*B*= Event of an odd number is chosen

*C*= Event of a prime number is chosen

*B*= {3, 5, 7, 9, 11, 13, 15} and

*n*(

*B*) = 7

*C*= {2, 3, 5, 7, 11, 13} and

*n*(

*C*) = 6

The
event when an odd number or a prime number is chosen is

*B*υ*C.**P*(

*B*υ

*C*) =

*P*(

*B*) +

*P*(

*C*) –

*P*(

*B*∩

*C*)

*B*∩

*C*= {3, 5, 7, 11, 13},

*n*(

*B*∩

*C*) = 5

$\begin{array}{l}P(B\cup C)\\ =\text{}P(B)+P(C)-P(B\text{}\cap \text{}C)\\ =\frac{n\left(B\right)}{n\left(S\right)}+\frac{n\left(C\right)}{n\left(S\right)}-\frac{n(B\cap C)}{n(S)}\\ =\frac{7}{14}+\frac{6}{14}-\frac{5}{14}\\ =\frac{8}{14}=\frac{4}{7}\\ \therefore \text{Theprobabilityofchoosingan}\\ \text{oddnumberoraprimenumber}=\frac{4}{7}.\end{array}$