**7.4 Probability of Independent Events**

**1.**

**In an experiment, if the outcomes of event A**

*do not influence*the outcomes of event B, then the two events are

**.**

*independent***2.**If

*A*and

*B*are two independent events, the probability for the occurrence of events

*A*and

*B*is

P(A
∩ B) = P(A) x P(B) |

**3.**The concept of the probability of two independent events can be expanded to three or more independent events. If

*A*,

*B*and

*C*are three independent events, the probability for the occurrence of events

*A*,

*B*and

*C*is

P(A
∩ B ∩ C) = P(A) x
P(B) x
P(C) |

**4.**

**A tree diagram can be constructed to show all the possible outcomes of an experiment.**

**Example:**

Fatimah, Emily and Rani are to take a written driving test. The probability that they pass the test are ½, ⅔ and ¾ respectively. Calculate the probability that

(a) only one of them passes the exam,

(b) at least two of them pass the exam,

(c) at least one of them passes the exam.

*Solution:*
Let

*P*= Pass and*F*= Fail
The tree diagram is as follows.

*P*(only one of them passes the exam)

=

*P*(*PFF*or*FPF*or*FFP*)
=

*P*(*PFF*) +*P*(*FPF*) +*P*(*FFP*)
$\begin{array}{l}=\frac{1}{24}+\frac{1}{12}+\frac{1}{8}\\ =\frac{1}{4}\end{array}$

**(b)**

*P*(at least two of them pass the exam)

=

*P*(*PPP*or*PPF*or*PFP*or*FPP*)
=

*P*(*PPP*) +*P*(*PPF*) +*P*(*PFP*) +*P*(*FPP*)
$\begin{array}{l}=\frac{1}{4}+\frac{1}{12}+\frac{1}{8}+\frac{1}{4}\\ =\frac{17}{24}\end{array}$

**(c)**

*P*(at least one of them passes the exam)

= 1 –

*P*(all of them fail)
= 1 –

*P*(*FFF*)