SurfStat.australia

Surfstat.australia: an online text in introductory Statistics

VARIATION AND PROBABILITY

INTRODUCTION

Discrete probability distributions
Consider an example with many possible outcomes: the age of person at his/her last birthday
Events:
A₁ : age < 20 years
A₂ : age 20 - 24 years and so on.
Age at last birthday (years)
<20 20-24 25-29 30-34 >34
Category Label A₁ A₂ A₃ A₄ A₅
Probability P(A₁) P(A₂) P(A₃) P(A₄) P(A₅)

In this example there are 5 categories (possible "events"). They do not overlap (there are no outcomes in common) so they are disjoint or mutually exclusive.

The probability that a person chosen at random will belong to category A₁ , say, is P(A₁).

The probability that a person belongs in category A₂ and category A₃ is zero

i.e. impossible because she/he cannot be aged 20-24 and 25-29

so P(A₂ and A₃) = 0.

The probability that a person belongs to category A₂ or A₃ is P(A₂) + P(A₃) (i.e. corresponding to the probability, of being in the age group 20-29 years).

In the above table the categories cover all possibilities, so they are said to be exhaustive. Each person must belong to one category so P(A₁ or A₂ or A₃ or A₄ or A₅) = 1. Because the events are also disjoint, this means that P(A₁) + P(A₂) + P(A₃) + P(A₄) + P(A₅) = 1. In general, if there are K events A₁,A₂,...,A_k that are disjoint and exhaustive, then P(A₁) + P(A₂) + ... + P(A_k) = 1.
For the above table, if a person is not aged 20-24 the he/she must be aged <20 or >24, so

P(not A₂) = P(A₁ or A₃ or A₄ or A₅)
= P(A₁) + P(A₃) + P(A₄) + P(A₅)
= 1 - P(A₂) because P(A_i) = 1

Complement
The complement of A_i is the event that A_i does not occur, but some other event occurs instead. Its probability is P(not A_i) = 1 - P(A_i)
Example - Reducing companies' discriminatory hiring practices
A company has come under pressure to eliminate discriminatory hiring practices (all its employees are overseas born women). Company officials have agreed with unions that during the next 5 years, 40% of their new employees will be men and 30% will be Australian born. 35% of new employees, though, will be overseas born women. What percentage of Australian born men are they committed to hire?

Men Women Total
Overseas 0.35
Australian ? 0.3
Total 0.4

Let us call the mutually exclusive and exhaustive events OM, OW, AM and AW.
P(W) = 1 - P(M) (complementary events)
P(W) = 1 - 0.4 = 0.6
P(W) = P(OW) + P(AW)
But P(W) = 0.6 and P(OW) = 0.35 so
P(AW) = P(W) - P(OW) = 0.6 - 0.35 = 0.25
P(A) = P(AM) + P(AW)
But P(A) = 0.3, P(AW) = 0.25
so P(AM) = P(A) - P(AW) = 0.3 - 0.25 = 0.05
i.e. 5% will be Australian born men.
Discrete Probability Distribution
This is a list of mutually exclusive and exhaustive outcomes of some process and the corresponding probabilities.
Example - 1 coin toss

Outcome: Head Tail
Probability: 1/2 1/2

Example - 1 fair die throw
There are 6 possible outcomes (i.e. #spots are 1,2,...,6)
Assume these are equally likely so each must have the probability of 1/6.

#spots: 1 2 3 4 5 6
Probability: 1/6 1/6 1/6 1/6 1/6 1/6

This is an example of a discrete uniform distribution.
Example - 3 coin tosses
Probability distribution for the number of heads obtained if 3 coins are tossed.
0 heads (TTT)
1 head (HTT, THT, TTH)
2 heads (HHT, HTH, THH)
3 heads (HHH)
There are 8 mutually exclusive and exhaustive outcomes.
Assume these are equally likely - i.e. each has a probability of 1/8
Then P(no heads) = P(TTT) = 1/8
P(one head) = P(HTT or THT or TTH)
= P(HTT) + P(THT) + P(TTH)
= 1/8 + 1/8 + 1/8 = 3/8
Similarly for 2 or 3 heads.
The probability distribution is

Number of heads: 0 1 2 3
Possibilities: 1/8 3/8 3/8 1/8

This is an example of a binomial probability distribution.

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	<20	20-24	25-29	30-34	>34
Category Label	A₁	A₂	A₃	A₄	A₅
Probability	P(A₁)	P(A₂)	P(A₃)	P(A₄)	P(A₅)