Discrete Random Variables

When you roll two dice, you usually care about one number: the sum. A random variable turns each outcome of an experiment into a number like that, so you can organize all the probabilities in one table or graph. This idea is the starting point for everything else in this unit: expected value, the binomial distribution, and the hypergeometric distribution.

Key ideas

Random variables

A random variable is a variable whose value is a number decided by the outcome of an experiment. We name it with a capital letter, like $X$, and write its possible values in lower case, like $x$.

• Roll two dice and let $X$ be the sum. The possible values are $x = 2, 3, \dots, 12$.
• Flip a coin $4$ times and let $X$ be the number of heads. The possible values are $x = 0, 1, 2, 3, 4$.

A random variable is discrete if you can list its values (usually whole numbers you get by counting). Measured quantities like time or mass are continuous; they come later, in continuous random variables.

$P(X = x)$ means “the probability that $X$ takes the value $x$”. For two dice, $P(X = 7) = \tfrac{6}{36}$.

Probability distributions

A probability distribution of $X$ matches each value $x$ with its probability $P(X = x)$. You can think of it as a mapping $x \to P(X = x)$, and it’s usually shown in a table:

$x$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P(X = x)$	$\tfrac{1}{36}$	$\tfrac{2}{36}$	$\tfrac{3}{36}$	$\tfrac{4}{36}$	$\tfrac{5}{36}$	$\tfrac{6}{36}$	$\tfrac{5}{36}$	$\tfrac{4}{36}$	$\tfrac{3}{36}$	$\tfrac{2}{36}$	$\tfrac{1}{36}$

Every probability distribution has two properties:

• each probability is between $0$ and $1$: $0 \le P(X = x) \le 1$;
• the probabilities add to $1$: $\sum P(X = x) = 1$.

(The symbol $\sum$, the Greek capital sigma, means “add up all of these”.)

Building a distribution

Without technology: list the sample space (with a table or tree diagram), find the value of $X$ for each outcome, then count. The sums table on probability and sample spaces gives the two-dice distribution above.

With technology: a spreadsheet makes the counting quick. List all $36$ pairs of dice in two columns, add them in a third column, then use COUNTIF to count how many times each sum appears and divide by $36$. You can also simulate: =RANDBETWEEN(1,6)+RANDBETWEEN(1,6) copied down $1000$ rows gives an experimental distribution that should come out close to the theoretical one.

Probability histograms

A probability histogram is a bar graph of a distribution:

• each value $x$ gets a bar of width $1$, centred on $x$ (so the bar for $7$ runs from $6.5$ to $7.5$);
• the bar’s height is $P(X = x)$;
• the bars touch, because the values are consecutive whole numbers.

Since each bar has width $1$, its area equals its probability, and the total area of all the bars is $1$.

The probability histogram for $X$, the sum of two dice. It’s symmetric about $x = 7$.

A frequency histogram looks similar but shows what actually happened in an experiment: bar heights are counts (or relative frequencies). It changes every time you repeat the experiment. A probability histogram shows what should happen in theory. With more and more trials, a relative frequency histogram gets closer and closer to the probability histogram.

The uniform distribution

A random variable has a uniform distribution if all its $n$ values are equally likely:

$$P(X = x) = \frac{1}{n} \quad \text{for each of the } n \text{ values}$$

Rolling one fair die is uniform with $n = 6$: each value has probability $\tfrac{1}{6}$. Its probability histogram is six bars of the same height, a flat “rectangle” shape.

To count the values from $a$ to $b$ (whole numbers), use $n = b - a + 1$. The numbers $1$ to $25$ have $n = 25$ values; the numbers $10$ to $99$ have $n = 90$.

Worked examples

Example 1: Using the two-dice distribution

Let $X$ be the sum of two fair dice. Use the table above to find $P(X \ge 9)$ and $P(4 \le X \le 6)$.

Solution. Add the probabilities of the values that fit.

$$P(X \ge 9) = \frac{4 + 3 + 2 + 1}{36} = \frac{10}{36} = \frac{5}{18}$$ $$P(4 \le X \le 6) = \frac{3 + 4 + 5}{36} = \frac{12}{36} = \frac{1}{3}$$

On the histogram, these are the total areas of the bars for $9$ to $12$, and for $4$ to $6$.

Example 2: Building a distribution by hand

A bag holds $3$ red and $2$ blue marbles. Two are drawn without replacement, and $X$ is the number of red marbles drawn. Make the probability distribution of $X$.

Solution. Use the tree diagram from independent and dependent events. The four paths are RR, RB, BR, BB.

• $X = 0$: only BB, so $P(X = 0) = \tfrac{2}{5} \times \tfrac{1}{4} = \tfrac{2}{20}$.
• $X = 1$: RB or BR, so $P(X = 1) = \tfrac{3}{5} \times \tfrac{2}{4} + \tfrac{2}{5} \times \tfrac{3}{4} = \tfrac{12}{20}$.
• $X = 2$: only RR, so $P(X = 2) = \tfrac{3}{5} \times \tfrac{2}{4} = \tfrac{6}{20}$.

$x$	$0$	$1$	$2$
$P(X = x)$	$\tfrac{1}{10}$	$\tfrac{3}{5}$	$\tfrac{3}{10}$

Check: $\tfrac{2}{20} + \tfrac{12}{20} + \tfrac{6}{20} = 1$. ✓

Example 3: A uniform distribution

At a school assembly, one ticket is drawn at random from tickets numbered $1$ to $25$. Let $X$ be the number drawn. Find $P(X = 13)$, $P(X \text{ is a multiple of } 4)$, and $P(X \gt 18)$.

Solution. All $25$ numbers are equally likely, so $X$ is uniform with $P(X = x) = \tfrac{1}{25}$.

$$P(X = 13) = \frac{1}{25}$$

The multiples of $4$ are $4, 8, 12, 16, 20, 24$: six values.

$$P(X \text{ is a multiple of } 4) = \frac{6}{25}$$

The values greater than $18$ are $19$ to $25$, which is $25 - 19 + 1 = 7$ values.

$$P(X \gt 18) = \frac{7}{25}$$

Example 4: Frequency versus probability

Two coins are flipped $40$ times, and $X$ is the number of heads each time. The results were: $0$ heads $12$ times, $1$ head $17$ times, $2$ heads $11$ times. Compare the relative frequencies with the theoretical probabilities.

Solution. The sample space is HH, HT, TH, TT, so the theoretical distribution is $P(X = 0) = \tfrac{1}{4}$, $P(X = 1) = \tfrac{1}{2}$, $P(X = 2) = \tfrac{1}{4}$.

$x$	$0$	$1$	$2$
Frequency	$12$	$17$	$11$
Relative frequency	$0.3$	$0.425$	$0.275$
Probability	$0.25$	$0.5$	$0.25$

The frequency histogram has the same general shape (tallest in the middle), but the heights don’t match exactly. That’s normal for only $40$ trials. With thousands of trials, the relative frequencies would get very close to $0.25$, $0.5$, $0.25$.

Common mistakes

Assuming every value of X is equally likely. The sums of two dice go from $2$ to $12$, but they are not uniform. Only use $P(X = x) = \tfrac{1}{n}$ when each value really is equally likely, like one fair die.

Forgetting to check that the probabilities add to 1. If your table adds to $0.95$ or $1.1$, a value is missing or a probability is wrong. Check every distribution you build.

Miscounting the values from a to b. From $10$ to $99$ there are $90$ whole numbers, not $89$. Use $b - a + 1$.

Drawing histogram bars in the wrong place. In a probability histogram, the bar for $x = 3$ goes from $2.5$ to $3.5$, centred on $3$, with height $P(X = 3)$. The bars touch.

Treating experimental results as the theoretical distribution. A frequency histogram from a small experiment is only an estimate. The probability histogram comes from the sample space, not from data.

Practice

1. (Warm-up) Is each random variable discrete or continuous?

• (a) the number of texts you get in a day
• (b) the time it takes to get to school
• (c) the number of heads in $10$ coin flips
• (d) the mass of your backpack

Solution

(a) Discrete. (b) Continuous. (c) Discrete. (d) Continuous.

2. (Warm-up) Is this a valid probability distribution? Explain.

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.1$	$0.3$	$0.4$	$0.3$

Solution

No. Each probability is between $0$ and $1$, but they add to $0.1 + 0.3 + 0.4 + 0.3 = 1.1$, not $1$.

3. (Warm-up) A spinner has $8$ equal sections numbered $1$ to $8$, and $X$ is the number spun. Write the distribution of $X$ in words, and find $P(X \ge 6)$.

Solution

$X$ is uniform: $P(X = x) = \tfrac{1}{8}$ for $x = 1, 2, \dots, 8$. The values $6, 7, 8$ give:

$$P(X \ge 6) = \frac{3}{8}$$

4. (Core) A random variable has this distribution. Find $k$, then find $P(X \ge 3)$.

$x$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$3k$	$4k$

Solution

The probabilities add to $1$: $k + 2k + 3k + 4k = 10k = 1$, so $k = 0.1$.

$$P(X \ge 3) = 3k + 4k = 0.3 + 0.4 = 0.7$$

5. (Core) A coin is flipped $4$ times, and $X$ is the number of heads. Make the probability distribution of $X$, and find $P(X \ge 3)$.

Solution

There are $2^4 = 16$ equally likely outcomes. Counting the outcomes with each number of heads (from a tree diagram, or row $4$ of Pascal’s triangle) gives $1, 4, 6, 4, 1$.

$x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$\tfrac{1}{16}$	$\tfrac{4}{16}$	$\tfrac{6}{16}$	$\tfrac{4}{16}$	$\tfrac{1}{16}$

$$P(X \ge 3) = \frac{4 + 1}{16} = \frac{5}{16}$$

6. (Core) Two dice are rolled, and $X$ is the difference between the larger and smaller numbers ($0$ for doubles). Make the probability distribution of $X$. Which value is most likely?

Solution

Make a $6 \times 6$ table of differences and count each value:

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$\tfrac{6}{36}$	$\tfrac{10}{36}$	$\tfrac{8}{36}$	$\tfrac{6}{36}$	$\tfrac{4}{36}$	$\tfrac{2}{36}$

Check: $6 + 10 + 8 + 6 + 4 + 2 = 36$. ✓ The most likely difference is $1$.

7. (Core) A die is rolled $60$ times with these results:

Face	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$8$	$12$	$9$	$11$	$7$	$13$

• (a) Find the relative frequency of each face, to $4$ decimal places.
• (b) Describe the probability histogram for a fair die, and how the frequency histogram differs from it.
• (c) What would you expect to see if the die were rolled $6000$ times?

Solution

(a) Divide each count by $60$: $0.1333$, $0.2000$, $0.1500$, $0.1833$, $0.1167$, $0.2167$.

(b) The probability histogram is uniform: six touching bars, each of height $\tfrac{1}{6} \approx 0.1667$. The relative frequency histogram is bumpy: some bars are above $0.1667$ (faces $2$, $4$, $6$) and some below (faces $1$, $3$, $5$).

(c) The relative frequencies would all be much closer to $0.1667$, so the frequency histogram would look almost flat, like the probability histogram.

8. (Challenge) You roll a die until you get a $6$, but you stop after $3$ rolls no matter what. Let $X$ be the number of rolls you make. Find the probability distribution of $X$.

Solution

• $X = 1$: the first roll is a $6$, so $P(X = 1) = \tfrac{1}{6}$.
• $X = 2$: not a $6$, then a $6$, so $P(X = 2) = \tfrac{5}{6} \times \tfrac{1}{6} = \tfrac{5}{36}$.
• $X = 3$: the first two rolls aren’t $6$s (the third roll happens whatever it shows), so $P(X = 3) = \tfrac{5}{6} \times \tfrac{5}{6} = \tfrac{25}{36}$.

$x$	$1$	$2$	$3$
$P(X = x)$	$\tfrac{6}{36}$	$\tfrac{5}{36}$	$\tfrac{25}{36}$

Check: $6 + 5 + 25 = 36$. ✓

9. (Challenge) A computer picks a two-digit whole number at random (from $10$ to $99$, all equally likely). Find the probability that its digits add up to $9$.

Solution

This is a uniform distribution with $n = 99 - 10 + 1 = 90$ values. The numbers whose digits add to $9$ are $18, 27, 36, 45, 54, 63, 72, 81, 90$: nine of them.

$$P = \frac{9}{90} = \frac{1}{10}$$