**1. Probability**

**Experiment:** an activity with a chance outcome and which is repeatable.

**Event:** a possible outcome or set of possible outcomes of the experiment. Typically denoted by a capital letter near the beginning of the alphabet: *A*,* B* etc.

**Probability of an event A: **denoted by P(

*A*). Measured on a scale between 0 and 1 inclusive, so that P(

*A*) = 0 indicates that

*A*is impossible and P(

*A*) = 1 indicates that

*A*is certain.

Intuitive idea: P(*A*) is the proportion of times *A* would occur if the experiment were repeated very many times.

*Example 1.1: Quality Control*

(a) Add the names of the events *E*1, *E*2, … , *E*8 to the left column.

Note : P(*E*1) + P(*E*2) + … + P(*E*8) = 1.

**Seven rules of probability**

**1. Special Addition Rule**

If several events *A**1**, A2**, … , Ak* are mutually exclusive (i.e. at most one of them can happen in a single experiment) then:

P(*A**1* or *A**2* or … or *A**k*) = P(*A**1**A**2* … *A**k*) = P(*A**1*) + P(*A**2*) + … + P(*A**k*)

**Example 1.1 (b)**

Find P(*B*), P(*C*) and P(*D*).

*Solution*

P(*B*) = P(*E*2 or *E*3 or* E*4) = P(*E*2) + P(*E*3) + P(*E*4)

=

P(*C*) = P(*E*1 or *E*2 or *E*3 or* E*4) = P(*E*1) + P(*E*2) + P(*E*3) + P(*E*4)

=

P(*D*) = P(*E*2 or *E*3 or* *… or *E*8) = P(*E*2) + P(*E*3) + … + P(*E*8)

=

**2. General Addition Rule**

For any two events *A*1 and *A*2:

P(*A*1 or *A*2) = P(*A*1*A*2)

= P(*A*1) + P(*A*2) – P(*A*1 and *A*2)

= P(*A*1) + P(*A*2) – P(*A*1 *A*2)

Note: “*A*1 or *A*2″ includes the possibility that both *A*1 and *A*2 occur.

**Example 1.1 (c)**

Find P(*C* or* D*)

*Solution*

P(*C* or *D*) = P(*C*) + P(*D*) – P(*C* and *D*)

P(*C* and* D*) = P(2 satisfactory __and__ 2 satisfactory)

= P(Exactly 2 satisfactory) = P(*A*) = 26/60.

P(*C* or *D*) = ,

not surprisingly since “*C* or *D*” is the event “either 2 or 2 satisfactory”, which must occur.

**3. Complements Rule**

P(Event happens) = 1 – P(Event doesn’t happen)

P(*A*) = 1 – P(*A**C*)

**Example 1.1 (d)**

Find P(*D*) using the Complements rule

*Solution*

P(*D*) = 1 – P(*D**C*)

= 1 – P(More than 2 items are satisfactory)

= 1 – P(*E*1) = 1 – .

**4. Special Multiplication Rule**

If two events *A*1 and *A*2 are *independent* then:

P(*A*1 and *A*2) = P(*A*1 *A*2) = P(*A*1) P(*A*2)

Independent events: knowing that *A*1 has occurred does not affect the probability that *A*2 has occurred.

**Example 1.1 (e)**

Check the given probabilities are correct.

*Solution*

Let* F**i* = event that the item from production Line *i* is satisfactory (*i* = 1, 2 or 3)

Then: P(*F*1) = 3/4; P(*F*2) = 2/3; P(*F*3) = 4/5.

P(*E*1) = P(*F*1 and *F*2 and *F*3) = P(*F*1) P(*F*2) P(*F*3) (since *F**i*‘s are independent)

= , as given above.

P(*E*2) = P(*F*1 and* F*2 and *F*3*C*) =, etc.

**Reliability of a system**

**General approach: **bottom-up analysis. Need to break down the system into subsystems just containing elements in series or just containing elements in parallel. Find the reliability of each of these subsystems and then repeat the process at the next level up.

**Series subsystem: **in the diagram *p**i* = probability that element *i* fails, so 1 –* p**i* = probability that it does not fail.

The system only works if all *n* elements work.

i.e. P(System does not fail) =

P(Element 1 doesn’t fail __and__ Element 2 doesn’t fail __and__ … __and__ Element *n* doesn’t fail)

= P(Element 1 doesn’t fail)P(Element 2 doesn’t fail) … P(Element *n* doesn’t fail)

[Special multiplication rule; independence of failures]

= (1-*p*1)(1-*p*2) … (1-*p**n*) =

**Parallel subsystem: **the subsystem only fails if all the elements fail.

i.e. P(System fails) = P(Element 1 fails __and__ Element 2 fails __and__ … __and__ Element *n* fails)

= P(Element 1 fails)P(Element 2 fails) … P(Element *n* fails)

[Independence of failures]

= *p*1*p*2 … *p**n* =

**Example 1.2(a)**

What is the probability that the system does not fail in the next year?

*Solution*

*Subsystem 1:*

P(Subsystem 1 doesn’t fail) = (1 – 0.05)(1 – 0.03) = 0.09215

P(Subsystem 1 fails) = 0.0785

*Subsystem 2:*

P(Subsystem 2 fails) = 0.0785 x 0.0785 = 0.006162

*Subsystem 3:*

P(Subsystem 3 fails) = 0.1 x 0.1 = 0.01

*System* *(summarised):*

P(System doesn’t fail) =

(1 – 0.02)(1 – 0.006162)(1 – 0.01) = 0.964

**Conditional probability: **the chance of event* A* occurring may be influenced by knowing that event *B* has occurred e.g. if *A* is the event “rain next Sunday” and *B* is the event “rain next Saturday”, the occurrence of *B* increase the chance of occurrence of *A*.

“The probability of event *A* given that event *B* has occurred” is denoted by P(*A* | *B*).

**5. General Multiplication Rule**

For any two events *A*1 and *A*2:

P(*A*1 and *A*2) = P(*A*1*A*2) =

**Example 1.2(b)**

Find P(System does not fail and component * does fail)

*Solution*

Let *B* = event that the system does not fail

Let *C* = event that component * does fail

We need to find P(*B* and *C*).

Now, P(*C*) = 0.1.

Also, P(*B* | *C*) = P(system does not fail given component * has failed);

now if component * has failed, Subsystem 3 has probability of failing of 0.1 instead of 0.01, so that the final reliability diagram becomes:

P(*B* | *C*) = (1 – 0.02) x

(1 – 6.162×10-3)(1 – 0.1) = 0.8766

P(*B* and *C*) = P(*B* | *C*) P(*C*) = 0.8766 x 0.1 = 0.08766

**Partition: **the events *A*1, *A*2, … , *A**k* form a partition if they are disjoint and their union includes all possible outcomes of the experiment.

**6. Theorem of Total Probability**

If *A*1, *A*2, … , *A**k* form a partition and *B* is any event:

P(*B*) = P(*B* | *A*1)P(*A*1) + P(*B* | *A*2)P(*A*2) + … + P(*B* | *A**k*)P(*A**k*)

**Example 1.1 (f)**

What is the probability that a randomly chosen item is satisfactory?

*Solution*

Let *G**i* = event that the selected item is from Line *i* (*i* = 1, 2, 3)

P(*G*1) = 0.4, P(*G*2) = 0.35, P(*G*3) = 0.25.

Let *S* be the event that the item is satisfactory.

Previously, we were told: P(*S* | *G*1) = 3/4; P(*S* | *G*2) = 2/3; P(*S* | *G*3) = 4/5

P(*S*) = P(*S* | *G*1)P(*G*1) + P(*S* | *G*2)P(*G*2) + P(*S* | *G*3)P(*G*3)

=

**7. Bayes Theorem**

If *A*1 and *A*2 are any two events such that P(*A*1) > 0 and P(*A*2) > 0, then

P(*A*1 | *A*2) =

Note: often the Theorem of Total Probability is used to evaluate P(*A*2).

**Example 1.2(c)**

Find P(Component * has failed | System has not failed)

*Solution*

Using the previously introduced notation:

P(*C* | *B*) = = = 0.091