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 E1, E2, … , E8 to the left column.

Note : P(E1) + P(E2) + … + P(E8) = 1.

Seven rules of probability

1. Special Addition Rule

If several events A1, A2, … , Ak are mutually exclusive (i.e. at most one of them can happen in a single experiment) then:

P(A1 or A2 or … or Ak) = P(A1A2Ak) = P(A1) + P(A2) + … + P(Ak)

Example 1.1 (b)

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

Solution

P(B) = P(E2 or E3 or E4) = P(E2) + P(E3) + P(E4)

=

P(C) = P(E1 or E2 or E3 or E4) = P(E1) + P(E2) + P(E3) + P(E4)

=

P(D) = P(E2 or E3 or … or E8) = P(E2) + P(E3) + … + P(E8)

=

2. General Addition Rule

For any two events A1 and A2:

P(A1 or A2) = P(A1A2)

= P(A1) + P(A2) – P(A1 and A2)

= P(A1) + P(A2) – P(A1 A2)

Note: “A1 or A2″ includes the possibility that both A1 and A2 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(AC)

Example 1.1 (d)

Find P(D) using the Complements rule

Solution

P(D) = 1 – P(DC)

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

= 1 – P(E1) = 1 – .

4. Special Multiplication Rule

If two events A1 and A2 are independent then:

P(A1 and A2) = P(A1 A2) = P(A1) P(A2)

Independent events: knowing that A1 has occurred does not affect the probability that A2 has occurred.

Example 1.1 (e)

Check the given probabilities are correct.

Solution

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

Then: P(F1) = 3/4; P(F2) = 2/3; P(F3) = 4/5.

P(E1) = P(F1 and F2 and F3) = P(F1) P(F2) P(F3) (since Fi‘s are independent)

= , as given above.

P(E2) = P(F1 and F2 and F3C) =, 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 pi = probability that element i fails, so 1 – pi = 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 andand 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-p1)(1-p2) … (1-pn) =

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

i.e. P(System fails) = P(Element 1 fails and Element 2 fails andand Element n fails)

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

[Independence of failures]

= p1p2 … pn =

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 A1 and A2:

P(A1 and A2) = P(A1A2) =

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 A1, A2, … , Ak form a partition if they are disjoint and their union includes all possible outcomes of the experiment.

6. Theorem of Total Probability

If A1, A2, … , Ak form a partition and B is any event:

P(B) = P(B | A1)P(A1) + P(B | A2)P(A2) + … + P(B | Ak)P(Ak)

Example 1.1 (f)

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

Solution

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

P(G1) = 0.4, P(G2) = 0.35, P(G3) = 0.25.

Let S be the event that the item is satisfactory.

Previously, we were told: P(S | G1) = 3/4; P(S | G2) = 2/3; P(S | G3) = 4/5

P(S) = P(S | G1)P(G1) + P(S | G2)P(G2) + P(S | G3)P(G3)

=

7. Bayes Theorem

If A1 and A2 are any two events such that P(A1) > 0 and P(A2) > 0, then

P(A1 | A2) =

Note: often the Theorem of Total Probability is used to evaluate P(A2).

Example 1.2(c)

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

Solution

Using the previously introduced notation:

P(C | B) = = = 0.091