**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) × 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:Solution:

Let

*P*= Pass and*F*= FailThe 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)

(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*)
$\begin{array}{l}=1-\frac{1}{24}\\ =\frac{23}{24}\end{array}$