Discrete Probability Distribution: Check It, Find the Missing Probability, Calculate the Mean

§ 01

What a discrete probability distribution is

Discrete probability distributions describe outcomes you can count and list — the number of defective parts in a batch of ten, the number of heads in three coin flips, the number of customers who call a helpline in an hour — each outcome paired with its own probability. That's different from the z table for the continuous case, where probability is area under a curve instead of a list of values, because a continuous variable like height or time can land anywhere in a range. A discrete distribution usually starts life as a frequency table, the same kind of table behind class width for the frequency table you start from, before its counts get converted into probabilities by dividing each count by the total. Once that conversion is done, the frequency table and the probability distribution describe the exact same data, just in two different units — counts on one, shares of 1 on the other.

§ 02

Is it a probability distribution? The two checks

To determine whether the distribution is a probability distribution, run two checks on the P(x) column and nothing else. Both checks are about the numbers in that column, not about how many outcomes there are or what they represent.

What a discrete probability distribution has to satisfy

A discrete probability distribution has to meet two conditions: every probability is between 0 and 1, and all the probabilities add up to exactly 1.

That's the whole test to determine whether the distribution is a discrete probability distribution — both conditions have to hold at once, and a distribution that fails either one doesn't qualify, however close it looks. A table that sums to 1 but has one negative entry fails. A table where every entry sits between 0 and 1 but the column sums to 0.9 or 1.1 fails too. Neither check substitutes for the other.

A valid distribution

Scroll the table sideways to see every column.

Every value sits between 0 and 1, and $0.20+$$0.30+$$0.30+$$0.20=$$0.20+0.30+0.30+0.20=1.00$ — both checks pass, so this distribution is valid.

Invalid: a negative value

Scroll the table sideways to see every column.

The column even sums to 1 exactly ($-0.10+$$0.50+$$0.60=$$-0.10+0.50+0.60=1.00$), but $P(1)=-0.10$ is negative, so the first check fails on its own. Summing to 1 never rescues a distribution that has a probability outside the 0-to-1 range.

Invalid: the column sums to 1.1

Scroll the table sideways to see every column.

Every individual value sits between 0 and 1 here, but $0.30+$$0.40+$$0.40=$$0.30+0.40+0.40=1.10$ — the second check fails, so this is not a valid discrete probability distribution either, even though nothing in the column looks obviously wrong on its own.

Both checks matter because they catch different kinds of mistakes. A negative or over-1 entry usually comes from a sign error or a miscount somewhere upstream; a column that doesn't sum to 1 usually means an outcome was left out, double-counted, or the counts were never converted into probabilities correctly by dividing by the total in the first place. Running both checks, in either order, takes a few seconds and catches almost every error before it reaches the mean or variance calculations further down this page.

§ 03

Determine the required value of the missing probability

To determine the required value of the missing probability, use the one rule every discrete distribution obeys: the whole column sums to 1. Add up the values you're given, subtract that total from 1, and whatever is left over is the missing probability — no other information about the variable is needed.

Finding the missing probability so the column sums to 1

Scroll the table sideways to see every column.

The four probabilities must sum to 1: $0.1+$$0.3+$$0.4=$$0.1+0.3+0.4=0.8$, so the missing value is $1-0.8=0.2$.

A second missing-probability table

Scroll the table sideways to see every column.

The three known probabilities add to $0.15+$$0.25+$$0.30=$$0.15+0.25+0.30=0.70$, so the missing value at $X=4$ is $1-0.70=0.30$. Check the full column once it's filled in: $0.15+$$0.30+$$0.25+$$0.30=$$0.15+0.30+0.25+0.30=1.00$, and every entry sits between 0 and 1 — a valid distribution.

§ 04

Calculate the mean for the discrete probability distribution

Calculate the mean for the discrete probability distribution shown here by multiplying each outcome by its own probability and adding the products — never by averaging the outcomes themselves. How to find the mean of a probability distribution comes down to one formula, $\mu=\sum x\cdot P(x)$: the mean for a probability distribution is always this probability-weighted sum.

Calculating the mean of a discrete probability distribution

The mean of a probability distribution is the sum of the last column: $\mu =$$\sum x\cdot$$P(x)=$$0+$$0.30+$$0.80+$$0.60=$$\mu = \sum x \cdot P(x) = 0+0.30+0.80+0.60 = 1.7$. This is the same expected-value calculation behind pricing an insurance product or a business decision tree; see accounting, finance and economics homework for that application. The expected value of a discrete probability distribution is this same number under a different name — μ and "expected value" describe identically the same calculation, just with different words attached.

A second worked mean

Scroll the table sideways to see every column.

$\mu =$$0+$$0.35+$$0.36+$$0.18=$$\mu=0+0.35+0.36+0.18=0.89$. This outcome, 0.89, is not one the variable can ever actually take — a mean rarely lands on a real, observable outcome, and that's expected, not an error.

A payout example: when the mean is zero

Scroll the table sideways to see every column.

$\mu =$$-2.50+$$0+$$1.50+$$1.00=$$\mu=-2.50+0+1.50+1.00=0$. An expected value of exactly 0 describes a fair game: over many repeats, the gains and the losses balance out exactly — the same reasoning an insurer or a business uses to price a decision where some outcomes are losses and others are gains.

§ 05

Variance and standard deviation of a discrete distribution

Variance measures how spread out a distribution's outcomes are around its mean — a small variance means the outcomes cluster near μ, a large one means they scatter far from it. The definitional formula is ${\sigma }^2=$$\sum (x-\mu {)}^2\cdot$$\sigma^2=\sum(x-\mu)^2\cdot P(x)$: for each outcome, find its distance from the mean, square that distance, weight it by the outcome's own probability, then add every term up. A faster shortcut gives the identical number without squaring each distance from $\mu$: ${\sigma }^2=$$\sum x^2\cdot$$P(x)-$$\sigma^2=\sum x^2\cdot P(x)-\mu^2$. The standard deviation is just the square root of the variance, $\sigma=\sqrt{\sigma^2}$, back in the same units as the outcomes themselves.

Scroll the table sideways to see every column.

The mean comes first, from the third column: $\mu =$$0+$$0.30+$$0.30+$$0.15=$$\mu=0+0.30+0.30+0.15=0.75$. Summing the last column gives the variance: ${\sigma }^2=$$0.28125+$$0.01875+$$0.234375+$$0.253125=$$\sigma^2=0.28125+0.01875+0.234375+0.253125=0.7875$, so $\sigma =$$\sqrt{0.7875}\approx$$\sigma=\sqrt{0.7875}\approx0.887$. The shortcut agrees: $\sum x^2\cdot$$P(x)=$$0+$$0.30+$$0.60+$$0.45=$$\sum x^2\cdot P(x)=0+0.30+0.60+0.45=1.35$, and $1.35-$${0.75}^2=$$1.35-$$0.5625=$$1.35-0.75^2=1.35-0.5625=0.7875$ — the same number, reached a different way, which is a useful check when the two methods disagree and one of them has an arithmetic slip in it.

Two distributions can share the same mean and still behave very differently, which is why variance gets its own calculation instead of riding along with the mean. A helpline that averages 0.75 complaints a day with almost no variance is predictable; one that averages 0.75 complaints a day but occasionally spikes to five is not, even though the two means are identical — the variance is what tells the two apart.

§ 06

Six worked tables, start to finish

These six items mix every skill above — checking validity, finding a missing probability, calculating a mean, and finding a variance — the way a real problem set does, rather than drilling one skill in isolation. Work each one before reading the answer underneath it, then check your own arithmetic against the line shown.

Composite sample written by GradeDraft for this page: the six tables below are not taken from a textbook or a client's assignment.

§ 07

Common mistakes

Three mistakes account for most wrong answers on this topic.

Dividing by n. The mean of a probability distribution is $\sum x\cdot P(x)$, not $(\sum x)\div n$ — you are not averaging the outcomes, you are weighting each one by how likely it is. Treating 0, 1, 2 and 3 as a plain four-item average gives 1.5, which is not the correct probability-weighted mean for any distribution unless every outcome happens to share the same probability.

Skipping the sum-to-1 check. A column of probabilities that looks reasonable at a glance can still fail it — 0.2, 0.3, 0.3 and 0.3 adds up to 1.1, not 1, which makes the whole distribution invalid even though every individual value sits inside the 0-to-1 range. Add the column before doing anything else with it.

Mixing up P(x) and x. In the products column, the outcome is what gets multiplied by its own probability, not the other way around — $x\cdot P(x)$, never $P(x)\cdot P(x)$ or $x\cdot x$. Students searching how to find mean of probability distribution tables most often lose the right column here, multiplying two probabilities together instead of an outcome by its probability, rather than making an arithmetic slip in the multiplication itself.

All three mistakes share one root cause: treating the P(x) column as decoration instead of as a weight. Once every outcome is multiplied by its own probability, and only its own probability, before anything gets added, the arithmetic that follows is ordinary addition and multiplication — nothing in the method itself is hard once the columns are lined up correctly.

§ 08

When the probability unit is due

A missing probability or a distribution's mean takes a few minutes once you know the two checks and the one formula. A full statistics problem set — several distributions, a hypothesis test, a dataset in SPSS or R — is where most students look for homework help by subject instead of working every step alone, and GradeDraft's statistics desk takes those on directly, the same way it handles a statistics problem set worked step by step.