**9.3 Probability of a Combined Event**

**9.3b Finding the Probability of Combined Events (a)**

*A*or*B*(b)*A*and*B*
$$\overline{)\begin{array}{l}\text{}P(A\text{or}B)=P(A\cup B)\text{}\\ \text{}=\frac{n(A\cup B)}{n(S)}\text{}\end{array}}$$

**2.**The probability of a combined event ‘

*A*and

*B*’ is given by the formula below.

**Example:**

The probabilities that two Form 5 students, Fiona and Wendy will pass the English oral test are
$\frac{1}{3}\text{and}\frac{2}{5}$
respectively. Calculate the probability that

**(a)**both Fiona and Wendy past the English oral test,

**(b)**both Fiona and Wendy fail the English oral test,

**(c)**either one of them passes the English oral test,

**(d)**at least one of them passes the English oral test.

Solution:Solution:

Let

*F*= Event that Fiona passes the English oral test

Therefore,

*F’*= Event that Fiona fails the English oral test

*W’*= Event that Wendy fails the English oral test

$\begin{array}{l}P\left(F\right)=\frac{1}{3},\text{}P\left(F\u2018\right)=\frac{2}{3}\\ P\left(W\right)=\frac{2}{5},\text{}P\left(W\u2018\right)=\frac{3}{5}\end{array}$

**(a)**

*P*(both Fiona and Wendy past the English oral test)

=

*P*(*F*∩*W*)**(b)**

*P*(both Fiona and Wendy fail the English oral test)

=

=

$\begin{array}{l}=\frac{2}{3}\times \frac{3}{5}\\ =\frac{2}{5}\end{array}$

*P*(*F’*∩*W’*)=

*P*(*F’**)*x*P*(*W’*)$\begin{array}{l}=\frac{2}{3}\times \frac{3}{5}\\ =\frac{2}{5}\end{array}$

**(c)**

*P*(either one of them passes the English oral test)

=

= (

*P*(*F*∩*W’*) +*P*(*F’*∩*W*)= (

*P*(*F**)*x*P*(*W’*)) + (*P*(*F’**)*x*P*(*W*))
$\begin{array}{l}=\left(\frac{1}{3}\times \frac{3}{5}\right)+\left(\frac{2}{3}\times \frac{2}{5}\right)\\ =\frac{7}{15}\end{array}$

**(d)**

*P*(at least one of them passes the English oral test)

= 1 –

*P*(Both of them fail) ← (concept of complement event)= 1 –

*P*(*F’**)*x*P*(*W’*)
$\begin{array}{l}=1-\frac{2}{5}\\ =\frac{3}{5}\end{array}$