4.7 Discrete random variables

Many random processes produce a number: the score on a die, the number of heads in three coin tosses, the amount of money won in a game, the number of insurance claims made in a year. A random variable gives that number a name, so that probabilities can be written and calculated with algebra instead of words. Its probability distribution lists every possible value with its probability, and its expected value describes the mean result over many repetitions.

Expected value is the basis of how casinos, lotteries and insurance companies set their prices. A company that sells thousands of insurance policies cannot predict which customers will make a claim, but it can predict its average cost per policy very accurately, and it sets its premiums from that average.

Random variables

Outcomes that are numbers

A random process (an experiment or trial) has outcomes that are not known in advance. When each outcome is described by a number, that number is a random variable. Random variables are written with capital letters such as $X$, $Y$ and $W$. The lower-case letter $x$ stands for one particular value that $X$ can take.

The word "discrete" means that the possible values can be listed, with gaps between them. A discrete random variable usually comes from counting (the number of heads, the number of defective items, the number of goals) or from a score or payout that takes a fixed set of values. Measured quantities such as time, mass and length can take any value in an interval; these are continuous random variables, which are treated with integration in 4.14 (HL).

The notation P(X = x)

The probability that $X$ takes the value $x$ is written $P(X = x)$. For example, if $X$ is the score on a fair die, then $P(X = 6) = \frac{1}{6}$ and $P(X \le 2) = \frac{2}{6} = \frac{1}{3}$. The statement inside the brackets is an event, so every rule of probability from 4.5 and 4.6 applies to it: complements, combined events and conditional probability.

Probability distributions

What a probability distribution is

A random variable is only useful once the probability of each of its values is known. Together, the values and their probabilities form the distribution of the random variable.

The most common form is a table with the values in the first row and their probabilities in the second. The distribution of the score $X$ on a fair die is

Every value here has the same probability. This simple pattern has its own name.

Building a distribution from a sample space

To find a distribution from first principles, list the equally likely outcomes in the sample space, work out the value of $X$ for each outcome, and count how many outcomes give each value. Dividing each count by the total number of outcomes gives the probabilities.

A sample space diagram (a grid of outcomes) does the same job when two dice are rolled. It is often the quickest way to see a pattern that leads to a formula for $P(X = x)$.

The distribution of the larger score rises steadily: high values are more likely, because there are more ways for at least one die to reach them. A bar chart, with the possible values on the horizontal axis and the probabilities as the heights of separate bars, shows the shape at a glance.

The conditions for a valid distribution

Since the values of $X$ cover every possible outcome, and no outcome gives two different values, the events $X = x$ are mutually exclusive and together make up the whole sample space. Their probabilities must therefore add up to 1. Each probability must also lie between 0 and 1, like any other probability.

The condition $\sum P(X = x) = 1$ gives an equation, so it can be used to find an unknown probability or constant in a distribution.

Distributions given by a formula

A distribution can also be given by a rule for $P(X = x)$ together with the list of values $x$ for which the rule applies. To use it, substitute each value of $x$ into the rule, which turns the formula into a table.

Sometimes the unknown appears in a quadratic, and the equation $\sum P(X = x) = 1$ has two solutions. Each solution must then be tested against the other condition, $0 \le P(X = x) \le 1$, and any solution that gives a negative probability (or one greater than 1) is rejected.

Probabilities of events involving X

Once the distribution is known, the probability of any event about $X$ is the sum of the probabilities of the values that make up the event. Inequalities need care with the end points, because $X$ takes only the listed values.

Conditional probabilities follow the rule $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ from 4.6, with events written in terms of $X$.

Expected value

The long-run mean of a random variable

Suppose a fair die is rolled 600 times. Each score is expected about $600 \times \frac{1}{6} = 100$ times, the expected number of occurrences from 4.5. The mean score of the 600 rolls would then be close to

The number of rolls cancels: each value is multiplied by its probability, and the products are added. The same reasoning works for any discrete random variable. If the process is repeated $N$ times, each value $x$ occurs about $N \times P(X = x)$ times, and the mean of the results is about

The expected value is a weighted mean of the possible values, with the probabilities as the weights. It is the same calculation as the mean of a frequency table, $\bar{x} = \frac{\sum f x}{\sum f}$, with relative frequencies replaced by probabilities. $E(X)$ is therefore a model of the long-run behaviour of a process: the relative frequencies of real data settle towards the probabilities, so the mean of real data settles towards $E(X)$.

Interpreting the expected value

The expected value need not be one of the possible values of $X$. No single roll of a die scores 3.5, and no set of three tosses gives 1.5 heads. $E(X)$ describes the average over many repetitions, not the result of one. For the same reason, $E(X)$ is not the most likely value: that is the value with the greatest probability (the mode), which for the larger score on two dice is 6, not 4.47.

Symmetric distributions

When a distribution is symmetric, its expected value is the value at the centre of symmetry. The distribution of the number of heads in three tosses has probabilities $\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$, symmetric about 1.5, so $E(X) = 1.5$ with no calculation. The sum of the scores on two fair dice is symmetric about 7, so its expected value is 7.

Using E(X) to find unknown probabilities

When a distribution has two unknown probabilities, the two facts $\sum P(X = x) = 1$ and a given value of $E(X)$ give a pair of simultaneous linear equations.

Expected value in games and decisions

Gain and fair games

In a game of chance a player usually pays a stake to play and may receive a prize. The amount the player actually ends up with is the gain:

A negative gain is a loss. If the stake is the same every time, then $E(\text{gain}) = E(\text{prize}) - \text{stake}$, because subtracting a fixed amount from every value of the prize subtracts it from their weighted mean as well.

A game with $E(\text{gain}) < 0$ favours the organiser, and one with $E(\text{gain}) > 0$ favours the player. Casino games and lotteries are designed so that the expected gain of a player is negative, which guarantees the organiser a profit over a large number of plays even though individual players sometimes win.

Finding the stake or prize that makes a game fair

Setting $E(\text{gain}) = 0$ gives an equation for an unknown stake or prize. Problems of this type are a direct application of expected value to the design of a game.

Raffles and lotteries

In a raffle, a fixed number of tickets is sold and a fixed set of prizes is given away. If every ticket is equally likely to win, the expected prize per ticket is the total prize money divided by the number of tickets.

Insurance

An insurance company charges a fixed premium and pays out if a claim is made. From the company's point of view, the payout is a random variable, and the expected payout per policy is its average cost. The premium must be greater than the expected payout for the company to make a profit, and the company also has to cover its running costs.

Every customer's expected gain from such a policy is negative, yet buying insurance can still be a sensible choice. Expected value describes the average over many repetitions, while one customer experiences a single year. A loss of $\$800$ may matter far more to one person than a certain cost of $\$90$. The expected value is a guide to the long run, and other considerations, such as the size of a possible loss, also matter in a real decision.

Exam-style questions on discrete random variables

Questions on this content combine three skills: building or completing a distribution, using it to find probabilities, and finding and interpreting an expected value. Many are set without a calculator, so fractions and exact working are common.

Self-check

• To do:

  I can explain what a discrete random variable is and list its possible values in a context.

• To do:

  I can build a probability distribution from a sample space, a sample space diagram or a tree diagram.

• To do:

  I can use $\sum P(X = x) = 1$ to find an unknown constant, and reject values that give impossible probabilities.

• To do:

  I can find probabilities such as $P(X > 3)$ and $P(X = 4 \mid X \ge 3)$ from a distribution.

• To do:

  I can calculate $E(X) = \sum x\,P(X = x)$ and explain why it need not be a possible value of $X$.

• To do:

  I can use $\sum P(X = x) = 1$ and a given $E(X)$ to find two unknown probabilities.

• To do:

  I can find the expected gain in a game and the stake or prize that makes it fair.

• To do:

  I can use expected value to analyse a raffle or an insurance policy.

It is given that $\mathrm{E}(X) = 1.4$.

1. (a)

   Find the value of $p$ and the value of $q$.

   [4]4 marks

   Markscheme (a)

   $p + 0.3 + q + 0.2 = 1$ (M1)

   $0 \times p + 1 \times 0.3 + 2q + 3 \times 0.2 = 1.4$ (M1)

   $q = 0.25$ A1

   $p = 0.25$ A1

2. (b)

   Find $\mathrm{P}(X \ge 2 \mid X \ge 1)$.

   [3]3 marks

   Markscheme (b)

   $\mathrm{P}(X \ge 1) = 0.75$

3. (c)

   Show that $\mathrm{P}(T = 3) = 0.25$.

   [3]3 marks

   Markscheme (c)

   recognition of the four outcomes $(0, 3), (3, 0), (1, 2), (2, 1)$

4. (d)

   Given that $T = 3$, find the probability that one of the spins scored 0.

   [2]2 marks

   Markscheme (d)

   $\dfrac{2 \times 0.25 \times 0.2}{0.25}$ (M1)

   $= 0.4$

5. (e)

   Let $F$ be the event "the first spin scores 1" and let $E$ be the event "$T = 3$". Determine whether the events $F$ and $E$ are independent. Justify your answer.

   [3]3 marks

(b) Both dice are at most $x$ in $x^2$ outcomes, and both are at most $x - 1$ in $(x - 1)^2$ outcomes. So the number of outcomes with larger score exactly $x$ is

Dividing by 36 gives $P(X = x) = \frac{2x - 1}{36}$.

(b) With $k = 0.1$, the probabilities are $0.1, 0.2, 0.3, 0.4$. Then

A quicker route uses the complement: $P(X \ge 2) = 1 - P(X = 1) = 1 - 0.1 = 0.9$.

(b) $X < 3$ means $X = 1$ or $X = 2$:

(b) The event "$X = 4$ and $X \ge 3$" is just $X = 4$, because 4 is at least 3. So

The $\frac{1}{36}$ cancels, so you can work with the counts: 7 outcomes out of the 32 in the reduced sample space.

The expected value is 2.4:

From (1), $a = 0.7 - b$. Substitute into (2):

Then $a = 0.7 - 0.4 = 0.3$.

Check: $0.2 + 0.6 + 1.2 + 0.4 = 2.4$. Both values lie between 0 and 1, so the distribution is valid.

Check: $30 + 48 + 12 = 90$.

(b) Weight each value by its probability and add:

(c) The prize is $\$5$ for each red ball, so the prize takes the values $0, 5, 10$ with the probabilities of $X = 0, 1, 2$. Its expected value is

For a fair game the stake equals the expected prize, so $c = 4$. The game is fair when it costs $\$4$ to play.

The expected payout is $\$28$ per policy.

(b) The expected profit per policy is $90 - 28 = \$62$. For 5000 policies, the expected profit is $5000 \times 62 = \$310\,000$.

(c) The premium $p$ must satisfy $p - 28 \ge 40$, so $p \ge 68$. The lowest premium is $\$68$.

(b) The distribution is $P(X = 1) = 0.4$, $P(X = 2) = 0.3$, $P(X = 3) = 0.2$, $P(X = 4) = 0.1$. Use the complement:

(c) Weight each value by its probability and add:

(d) The prize takes the values $3, 6, 9, 12$ with probabilities $0.4, 0.3, 0.2, 0.1$:

The game is fair when $E(\text{gain}) = 6 - c = 0$, so $c = 6$.

(e) With $c = 7$, the expected gain per game is $6 - 7 = -\$1$. Over 20 games the expected total gain is $20 \times (-1) = -\$20$, an expected loss of $\$20$.

(c) He rolls exactly twice when the first roll is not a six and the second is: $P(X = 2) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$.

Check: $\frac{6 + 5 + 25}{36} = 1$.

(d) Use $E(X) = \sum x\,P(X = x)$:

(e) The expected number of occurrences is $np$: $72 \times \frac{1}{6} = 12$ games.

recognition that the events $X \ge 2$ and $X \ge 1$ both occur exactly when $X \ge 2$ (M1)

$\dfrac{0.45}{0.75} = 0.6$ A1

The spinner is spun twice. The scores on the two spins are independent. Let $T$ be the total of the two scores.

$2 \times 0.25 \times 0.2 + 2 \times 0.3 \times 0.25$ A1

$= 0.1 + 0.15$ A1

$= 0.25$ AG

Markscheme (e)

[4]4 marks

[2]2 marks