2.04a - Finite discrete probability distributions

2.04a - Finite discrete probability distributions

Use a table or formula to describe a finite discrete random variable, then turn a verbal event into the values you need to add. You will check that a proposed distribution is valid and combine independent observations without losing an order or counting it twice.

A random variable assigns numbers to outcomes

Can two inspections be more likely to find two faults in total than one? Start with one inspection: its fault count is 0, 1 or 2, with probabilities 0.20, 0.50 and 0.30. The most likely single count is 1. That does not yet tell you the most likely total.

You already know

Probabilities lie between 0 and 1. Add probabilities of mutually exclusive events; multiply probabilities of independent events. A complement has probability one minus the event's probability.

The inspection is random, but the rule assigning a number to its outcome is precise.

Finite discrete random variable

A random variable assigns a numerical value to each outcome of a random experiment. A finite discrete random variable has a finite set of possible values, called its support. Write a capital letter such as XX for the variable and a lower-case xx for a value. For example, XX is the fault count; x=2x=2 is one possible value. A measurement allowed to take any real value in an interval is not finite discrete.

Here is the complete distribution. Its three events exhaust the possibilities and do not overlap.

xx012
P(X=x)\mathrm P(X=x)0.200.500.30

Probabilities at fault counts 0, 1 and 2; no probability between the spikes

The spike at 1 has height 0.50. There is no spike at 1.5: P(X=1.5)=0\mathrm P(X=1.5)=0. A probability table predicts proportions over repetitions; it does not guarantee exact frequencies in a small sample.

Worked example 1: including the endpoint

Question. For this fault-count distribution, find the probability of at most one fault.

Plan. Translate “at most” into an inclusive inequality and add the allowed entries.

P(X≤1)=P(X=0)+P(X=1)include both counts=0.20+0.50=0.70disjoint events\begin{aligned} \mathrm P(X\leq1)&=\mathrm P(X=0)+\mathrm P(X=1) && \text{include both counts} \\ &=0.20+0.50=0.70 && \text{disjoint events} \end{aligned}

Why this step? At most one permits one fault as well as none.

Answer. The probability is 0.70.

Check. Its complement is two faults, with probability 0.30; the probabilities total 1.

Change only whether the endpoint is included. That small wording change removes a whole spike.

Worked example 2: excluding the endpoint

Question. For the same distribution, find the probability of fewer than one fault.

Plan. List the support values strictly below 1.

{x:x<1}={0}within the stated supportP(X<1)=P(X=0)=0.20only one allowed count\begin{aligned} \{x:x<1\}&=\{0\} && \text{within the stated support} \\ \mathrm P(X<1)&=\mathrm P(X=0)=0.20 && \text{only one allowed count} \end{aligned}

Why this step? “Fewer than” excludes 1, whose probability is 0.50.

Answer. The probability is 0.20.

Check. The difference from 0.70 is exactly P(X=1)=0.50\mathrm P(X=1)=0.50. Also P(X<1.5)=0.70\mathrm P(X<1.5)=0.70, since the allowed values below 1.5 are 0 and 1.

For an interval, make the same value-by-value decision at both ends.

Watch out

Wrong: P(X<1)=P(X≤1)\mathrm P(X<1)=\mathrm P(X\leq1). For a discrete variable, the difference is P(X=1)\mathrm P(X=1), which can be positive. Read the inequality before touching the calculator.

Try a different complete table; the endpoint is still the key decision.

Probabilities must sum to one

A formula can replace a table, but it needs a stated support and must assign valid probabilities to every value in it.

A valid finite probability distribution

For a finite support SS,

0≤P(X=x)≤1(x∈S),∑x∈SP(X=x)=10\leq\mathrm P(X=x)\leq1\quad(x\in S),\qquad \sum_{x\in S}\mathrm P(X=x)=1

The probability is zero outside the support. Not in the formula sheet: learn it.

Use the sum condition to find an unknown constant, then check the individual probabilities. The sum alone cannot rule out negative entries.

Worked example 3: normalising a rule

Question. A variable has P(X=x)=kx\mathrm P(X=x)=kx for x=1,2,3,4x=1,2,3,4, and zero otherwise. Find kk and P(X≥3)\mathrm P(X\geq3).

Plan. Sum the rule over its complete support, then add the entries for 3 and 4.

k(1+2+3+4)=1total probability10k=1⇒k=110solveP(X≥3)=310+410=710allowed values only\begin{aligned} k(1+2+3+4)&=1 && \text{total probability} \\ 10k&=1\quad\Rightarrow\quad k=\frac1{10} && \text{solve} \\ \mathrm P(X\geq3)&=\frac3{10}+\frac4{10}=\frac7{10} && \text{allowed values only} \end{aligned}

Why this step? The support says exactly which terms belong in the normalising sum.

Answer. k=1/10k=1/10 and the requested probability is 0.70.

Check. The entries 0.1, 0.2, 0.3 and 0.4 are non-negative and total 1; P(X=5)=0\mathrm P(X=5)=0, not 5k5k.

Keep the rule and the event, but extend the support by one value. The constant must change because the total probability is still 1.

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