VARIATION AND PROBABILITY

In general, if events X and Y are not mutually exclusive then

Example - Fruit in 2 districts

A certain kind of fruit is grown in 2 districts, A and B. Both areas sometimes get fruitfly.

Suppose the probabilities are

What is the probability that one or other (or both) districts are infected at a given time?

Alternatively, consider mutually exclusive parts

A or B consists of 3 mutually exclusive parts: A only, B only, A and B.

P(A or B) = P(A only) + P(B only) + P(A and B)

Events X and Y are said to be independent if the probability that X occurs is not affected by whether or not Y has occurred. It can be shown that this implies:

P(X and Y) = P(X)P(Y)

This is called the multiplication rule for independent events.

Example - 2 security guards and their pagers

There are two security guards for a large establishment. Each carries a pager activated by detectors in the buildings. Guard 1 is conscientious and is within earshot of his pager 80% of the time. Guard 2 is not as reliable and can only respond to the pager 50% of the time.

If the guards independently report any alert to the police or fire brigade, what is the probability that at least one will report an alert?

Let X be the event that Guard 1 reports the alert. P(X) = 0.8

Let Y be the event that Guard 2 reports the alert. P(Y) = 0.5

Are events X and Y mutually exclusive? - No, both can report.

Are X and Y independent? - assume Yes.

P(at least one reports)

But P(X and Y) = P(X) P(Y) (independent.)

so P(X or Y) = 0.8 + 0.5 - 0.4 = 0.9

Thus although Y is only reliable 50% of the time, employing him does increase the probability of reporting an alert.

Tree diagrams are helpful in organising calculations that involve several stages. Each segment in the tree is one stage of the problem and the probabilities on the branches from each point must sum to 1. The probability of reaching the end of any complete path is the product of the probabilities written on its branches.

Example - Boys and girls in a family of 3

Tree model for the order of boys (B) and girls (G) in a family of size 3.

Figure 1

Each path represents an outcome (family of 3). There are 8 outcomes. If you assume these are equally likely then the probability of each is 1/8.

Another way of calculating this is to assume that at each birth

Then e.g. P(BGB) = 1/2 x 1/2 x 1/2 = 1/8 = 0.125

i.e. assuming sex is independent of previous births and multiplying probabilities along the branches of the tree.

Tree models are useful for analysing any "step-by-step" process.

(In fact P(B) = P(G) = 1/2 is not quite correct.)

Example - Gender in human populations

In human populations about 52% of births are boys and 48% are girls. So a more realistic model is to use

Figure 2 - More realistic model

e.g. P(BGB) = 0.52 x 0.48 x 0.52 = 0.13

An event is any subset of outcomes.

Calculate probabilities for the following events using the "realistic" model.

	GGG		GGG		GGG		GGG
	GGB		GGB		GGB		GGB
	GBG		GBG		GBG		GBG
	GBB		GBB		GBB		GBB
	BGG		BGG		BGG		BGG
	BGB		BGB		BGB		BGB
	BBG		BBG		BBG		BBG
	BBB		BBB		BBB		BBB
	C		D		E		F

P(C) = P(GGG) + P(BBB) = 0.11 + 0.14 = 0.25

P(D) = 0.11 + 3 x 0.12 = 0.47

P(E) = P(C and D) = P(GGG) = 0.11

P(F) = P(C or D) = 0.11 + 3 x 0.12 + 0.14 = 0.61

Events C and D are not mutually exclusive (disjoint) because outcome GGG is in both. Both C and D can happen at once. Therefore P(C or D) = P(F) is not the same as P(C)+P(D), because this would count the common outcome (GGG) twice.

[compare this with the addition rule for probabilities of mutually exclusive events].

Instead, use the more general rule for P(C or D)