Random Variables and Probability Distributions: Check the Table, Find the Mean

A probability table may look like a small collection of fractions. In fact, it tells a complete story about what could happen and how likely each result is.

There are two separate jobs in most questions. First, decide whether the table tells a possible story. Then find its mean, which describes the balance of those possible results over many repetitions.

If you jump straight to the mean, an impossible table can still produce a perfectly neat number. That neat number is not a valid answer.

Let’s make that routine feel natural rather than mechanical.

What Is a Random Variable?

A random variable assigns a number to the result of a chance experiment. We usually call it X.

For example, toss two fair coins and let X be the number of heads. The possible outcomes and their values are:

Coin outcome	Value of X
TT	0
HT	1
TH	1
HH	2

The outcomes HT and TH are different coin outcomes, but both give the same value of X. That distinction matters. The random variable is not the entire outcome; it is the number we have chosen to record from that outcome.

Here X can take only 0, 1, or 2. These separate, countable possibilities make it a discrete random variable. The number of customers who arrive, defective items in a sample, or correct answers on a quiz are other examples.

Build the Distribution From Outcomes

With two fair coins, the four outcomes in the table above are equally likely. Count how many outcomes produce each value of X:

X, the number of heads	Outcomes	P(X = x)
0	TT	1/4
1	HT, TH	2/4 = 1/2
2	HH	1/4

This is the probability distribution of X: it pairs every possible value with the probability of getting that value.

Notice what happens when we add the probabilities:

1/4 + 1/2 + 1/4 = 1

The total is one because something in our complete list must happen.

For unequal outcomes, do not simply count cases unless they are equally likely. Work out the probability of each event, combine the events that give the same value of X, and then check the resulting table.

The Three Checks Before You Calculate a Mean

For the finite discrete tables in this guide, use three checks in this order.

1. Meaning: Does each listed x fit the experiment? Have you included every possible value, or accounted for a value that has zero probability?
2. Individual probabilities: Is every P(X = x) between 0 and 1, inclusive?
3. Total: Do all the probabilities add to exactly 1?

The first check makes sure the table represents the stated experiment. The next two are the numerical tests for a valid discrete probability distribution.

Proposed probabilities	Total	Verdict
0.2, 0.5, 0.3	1	Valid if the values of X are complete and appropriate
0.2, 0.5, 0.4	1.1	Invalid: the total is too large
0.2, 0.5, 0.1	0.8	Incomplete or invalid as a complete table
-0.1, 0.6, 0.5	1	Invalid: a probability cannot be negative
0, 0.4, 0.6	1	Valid: an impossible listed value may have probability zero

The negative-probability row is a useful warning. Adding to one is necessary, but it is not enough by itself.

If an otherwise sensible table totals 0.8, you cannot quietly turn its entries into percentages of 0.8. A missing 0.2 might belong to an omitted possible value, or the given figures might be wrong. Read the question before repairing anything.

Find a Missing Probability First

Suppose X can be 0, 1, 2, or 3, with this table:

X	0	1	2	3
P(X = x)	0.15	0.35	k	0.20

The total must be one:

0.15 + 0.35 + k + 0.20 = 1
0.70 + k = 1
k = 0.30

Now check that 0.30 is between zero and one, and that the four entries total one. Only now is the table ready for a mean calculation.

Sometimes the question gives several multiples of an unknown constant. For example:

X	-1	0	2	4
P(X = x)	k	2k	3k	4k

Add the probabilities, not the values of X:

k + 2k + 3k + 4k = 1
10k = 1
k = 1/10

The probabilities are 1/10, 2/10, 3/10, and 4/10. All are valid. The negative value X = -1 is no problem: a random variable can describe a loss, temperature change, or any other numerical result. It is a negative probability that would be impossible.

Mean or Expected Value: A Weighted Average

Once the table is valid, multiply each possible value by its probability and add the products:

E(X) = sum of (each value * its matching probability)

The probabilities are the weights. A value that occurs more often contributes more to the average.

Return to the two-coin table:

X	P(X = x)	x * P(X = x)
0	1/4	0
1	1/2	1/2
2	1/4	1/2
Total	1	1

Therefore:

E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1

If you repeatedly toss two fair coins, the average number of heads per pair of tosses tends toward one. This does not promise exactly one head on the next pair. The next result can still be zero, one, or two.

Complete the Earlier k Example

We found k = 1/10 for values -1, 0, 2, and 4. Now calculate the mean:

X	P(X = x)	x * P(X = x)
-1	1/10	-1/10
0	2/10	0
2	3/10	6/10
4	4/10	16/10
Total	1	21/10

So E(X) = 21/10 = 2.1.

The negative value lowers the mean. Leaving out its minus sign would produce 2.3, a plausible-looking but incorrect answer.

Why a Mean Need Not Be a Possible Outcome

Suppose a delivery desk receives 0, 1, or 2 urgent parcels in a chosen hour with probabilities 0.2, 0.5, and 0.3.

E(X) = 0(0.2) + 1(0.5) + 2(0.3)
     = 0 + 0.5 + 0.6
     = 1.1

You will never see 1.1 parcels in one hour. The number is an average across many such hours, not a prediction that a single hour can take a fractional value.

Nor must the mean be the most likely outcome. Here 1 is most likely, while the mean is 1.1 because the possibility of 2 pulls the balance upward.

An even sharper example is X = 0, 1, or 10 with probabilities 0.7, 0.2, and 0.1. Zero is by far the most likely result, but:

E(X) = 0(0.7) + 1(0.2) + 10(0.1) = 1.2

The rare value 10 still contributes a whole 1 to the mean. This is why a weighted average is more informative than simply pointing at the highest probability.

Turn an Experiment Into a Table: Two Dice

Let X be the number of sixes when two fair dice are rolled. Its possible values are 0, 1, and 2.

There are 36 equally likely ordered outcomes.

• No six: 5 x 5 = 25 outcomes.
• Exactly one six: 1 x 5 + 5 x 1 = 10 outcomes.
• Two sixes: 1 x 1 = 1 outcome.

The distribution and mean are:

X	Number of outcomes	P(X = x)	x * P(X = x)
0	25	25/36	0
1	10	10/36	10/36
2	1	1/36	2/36
Total	36	36/36	12/36

Hence E(X) = 12/36 = 1/3. If you roll two dice many times, you average one six for every three pairs of dice rolled. It does not mean that one-third of a six appears in a single roll.

This example also gives a quick reasonableness check: the mean must lie between the smallest and largest possible values, here 0 and 2.

Events From a Distribution: Do Not Mix Up Values and Probabilities

Once a table is valid, it can answer questions other than the mean. Use the values of X to decide which columns belong to the event, then add their probabilities.

For the number of heads in two coin tosses:

P(X >= 1) = P(X = 1) + P(X = 2)
          = 1/2 + 1/4
          = 3/4

And:

P(X < 2) = P(X = 0) + P(X = 1)
         = 1/4 + 1/2
         = 3/4

The expressions happen to have the same answer in this example, but they describe different events.

For the earlier four-value table with X = -1, 0, 2, 4:

P(X > 0) = P(X = 2) + P(X = 4)
         = 3/10 + 4/10
         = 7/10

Do not add 2 + 4. The values identify the columns; the probabilities measure the chance.

When the Question Gives the Mean Instead

Sometimes a table has an unknown probability and a stated mean. Use the probability total and the mean equation together.

Suppose:

X	0	1	3
P(X = x)	1/4	k	3/4 - k

The probability total is already one for every k, so that condition alone cannot find k. We still need 0 <= k <= 3/4 to keep the entries valid.

If the mean is 3/2, write:

0(1/4) + 1(k) + 3(3/4 - k) = 3/2
k + 9/4 - 3k = 3/2
9/4 - 2k = 6/4
2k = 3/4
k = 3/8

Then 3/4 - k = 3/8. The probabilities are 1/4, 3/8, and 3/8. Check them once more: they total one, lie within the allowed range, and produce the stated mean.

Mean of a Changed Variable

Suppose X counts correct answers, and a score is defined by Y = 3X + 5. Every possible value of X is tripled and then increased by five. Its mean changes in the same way:

E(Y) = 3E(X) + 5

For the two-coin distribution, E(X) = 1, so E(Y) = 8.

Check it directly: X = 0, 1, 2 gives Y = 5, 8, 11, with the same probabilities 1/4, 1/2, 1/4.

E(Y) = 5(1/4) + 8(1/2) + 11(1/4)
     = 5/4 + 4 + 11/4
     = 8

This shortcut is helpful in scoring and payoff questions. If X is a prize and every player pays a fixed entry fee of 5, the mean net payoff is E(X) - 5, not E(X).

A Fast Checklist for Any Table Question

Before you finish, ask yourself:

1. Have I said clearly what X measures and listed its possible values?
2. If I built the table from outcomes, did I avoid treating unequal outcomes as equally likely?
3. Is every probability between zero and one?
4. Does the complete probability column add to one?
5. Did I find missing probabilities before calculating the mean?
6. Did I multiply each value by its matching probability, including negative values and zero?
7. Does the mean lie between the smallest and largest possible values?
8. Did I explain the answer as a long-run average, not a guaranteed next result?

The range check can catch many mistakes, but it cannot prove a mean is correct. A wrong calculation may still land inside the range. The full working matters.

Try These Yourself

Pause before reading the answers. For each question, use the same order: validity, any missing probability, then the requested calculation.

1. X = 0, 1, 2 has probabilities 1/5, 1/2, 3/10. Is the table valid? Find E(X).
2. X = 0, 1, 2 has probabilities 0.20, 0.35, 0.55. Is it valid? Explain without calculating a mean.
3. X = -2, 0, 3 has probabilities k, 2k, 2k. Find k and E(X).
4. X is the number of heads in two fair coin tosses. Find P(X >= 1) and E(X).
5. X = 1, 3, 5 has probabilities 1/4, 1/2, 1/4. Find the mean. Is it necessarily a value obtained on the next trial?
6. X = 0, 2, 4 has probabilities 0.2, 0.5, 0.3. Find P(X >= 2) and E(X).
7. X = 0, 1, 2 has probabilities 0.25, k, 0.40. Find k and the mean.
8. A game has net payoff X = -5 with probability 0.8, or X = 10 with probability 0.2. Find the mean net payoff and interpret its sign.
9. X = 0, 2, 4 has probabilities k, 1/2, 1/2 - k. If E(X) = 2, find k.
10. X = 0, 1, 2 has probabilities 0.1, 0.6, 0.3. If Y = 2X - 3, find E(Y).

Answers With Short Checks

1. Valid: the probabilities total one. E(X) = 0 + 1/2 + 6/10 = 11/10 = 1.1.
2. Invalid: 0.20 + 0.35 + 0.55 = 1.10. Do not report a mean for this table.
3. 5k = 1, so k = 1/5. The mean is -2/5 + 0 + 6/5 = 4/5.
4. P(X >= 1) = 3/4 and E(X) = 1.
5. E(X) = 1/4 + 3/2 + 5/4 = 3. This happens to be a possible value, but the next trial is not guaranteed to give 3.
6. P(X >= 2) = 0.5 + 0.3 = 0.8; E(X) = 0 + 1 + 1.2 = 2.2.
7. k = 1 - 0.25 - 0.40 = 0.35; E(X) = 0 + 0.35 + 0.80 = 1.15.
8. E(X) = -5(0.8) + 10(0.2) = -2. The long-run average is a loss of 2 per play, although any one play gives either a loss of 5 or a gain of 10.
9. E(X) = 2(1/2) + 4(1/2 - k) = 3 - 4k. Thus k = 1/4, making the last probability 1/4.
10. E(X) = 0 + 0.6 + 0.6 = 1.2; hence E(Y) = 2(1.2) - 3 = -0.6.

If one of these felt difficult, identify which stage caused the trouble. Was it describing X, checking the probability total, choosing the right columns for an event, or forming the weighted mean? Fixing the stage is more useful than memorising another answer.

For an earlier step involving repeated draws, see probability with and without replacement. It explains how the chances change before you build a distribution.

Further Reading

• NIOS, Probability, Lesson 19 develops random variables, distributions, the validity checks, and the mean.
• OpenStax, Discrete Probability Distributions explains how to read and check a discrete table.
• OpenStax, Mean or Expected Value explains the weighted mean and its long-run interpretation.

Frequently Asked Questions

What is the difference between an outcome and a value of a random variable?

An outcome is what actually happens, such as HT in two coin tosses. A random variable assigns that outcome a number. If X counts heads, both HT and TH give X = 1.

Does every probability have to be strictly greater than zero?

No. Zero is allowed. It means that the listed value cannot occur under the stated experiment. Each probability must be at least zero and at most one.

Is a table valid just because its probabilities add to one?

No. A negative entry, an entry greater than one, or a list that misrepresents the experiment can still make the table invalid. Check the entries and what X means as well as the total.

What if the probabilities add to less than one?

As a complete distribution, the table is not valid. Look for an omitted possible value or a missing probability specified by the question. Do not calculate the mean until the table is complete.

What if the probabilities add to more than one?

The table cannot describe a probability distribution as written. Recheck arithmetic and whether overlapping events were counted twice.

Can a random variable have a negative value?

Yes. A net loss or a change below a reference point can be negative. Its probability must still be between zero and one.

Why do we multiply each value by its probability?

The probability tells us how much weight to give that value in the long-run average. Adding the weighted contributions gives E(X).

Can the mean be a fraction when X counts whole objects?

Yes. An average over many repetitions can be fractional even when every single observation is a whole number. 1.1 parcels per hour does not describe a fractional parcel in one hour.

Is the mean always the most likely value?

No. The most likely value has the largest individual probability. The mean uses all values and probabilities, so a rare large value can pull it away from the most likely value.

Is expected value what will happen next?

No. Expected value describes a long-run average under the given probabilities. The next trial can produce any of the permitted values.

How do I find P(X > 1) from a table?

Select only the columns whose X values are greater than one, then add their probabilities. Do not add the X values themselves.

Should I find k from the total or the mean?

Use the probability total first. If that equation does not determine k because the table already totals one for every k, use the given mean or another stated condition. Finally check that every resulting probability is valid.