OMPT drill — Probability & statistics

OMPT random variables practice: lesson and 10 problems

Expected value, variance, and reading a probability distribution table. The arithmetic is light; the notation is what needs rehearsal. Write the formula down before plugging in numbers; on this strand the setup is the whole battle.

Tested inOMPT-E

OMPT random variables: the lesson

What the OMPT tests in random variables

Random variables cover two of the twelve OMPT-E chapters, "Random variables" and "Probability distributions", and appear on no other variant. A random variable attaches a number to each outcome of a chance experiment: the sum of two dice, the number of heads in ten flips, the payout of a raffle ticket. Its behaviour is summarised in a distribution table, each possible value beside its probability, with the probabilities summing to exactly 1. Every question in the first chapter is answered by reading that table correctly: find a missing probability, compute the expected value $E (X)$ , compute the variance and standard deviation, and track what happens to them when the variable is rescaled.

The second chapter adds the binomial distribution: a fixed number of independent trials with the same success probability, and questions of the form "probability of exactly $k$ successes" or "at least one success". The arithmetic here is light. What needs rehearsal is the notation, because $E (X^{2})$ and $(E (X))^{2}$ look like twins and differ by exactly the variance. Misreading one for the other is the single most common slip I mark on this chapter.

Notation and answer format the OMPT expects

$E (X)$ or $E [X]$ for the expected value, sometimes $μ$ . $Var (X)$ for the variance, $σ$ for the standard deviation, which is $Var (X)$ .
A binomial variable is written $X \sim B (n, p)$ : $n$ trials, success probability $p$ . Its mean is $n p$ and its variance $n p (1 - p)$ .
$P (X = k)$ , $P (X \leq k)$ and $P (X \geq k)$ mean exactly what they say; "at most 2" is $\leq 2$ , "fewer than 2" is $\leq 1$ .
Expected values are typed as decimals or fractions (3.5, or 7/2). Probabilities from a binomial are rounded to the number of decimals the question names.

Worked example 1: a full table workout

$X$ takes the value 1 with probability 0.2, the value 2 with probability 0.5, and the value 3 with probability 0.3. Find $E (X)$ , $Var (X)$ and the standard deviation.
First confirm the table is legal: $0.2 + 0.5 + 0.3 = 1$ . Good.
Expected value: $E (X) = 1 (0.2) + 2 (0.5) + 3 (0.3) = 0.2 + 1.0 + 0.9 = 2.1$ .
For the variance, first $E (X^{2})$ , squaring the values and not the probabilities: $E (X^{2}) = 1 (0.2) + 4 (0.5) + 9 (0.3) = 0.2 + 2.0 + 2.7 = 4.9$ .
Then $Var (X) = 4.9 - (2.1)^{2} = 4.9 - 4.41 = 0.49$ , and the standard deviation is $0.49 = 0.7$ .
Plausibility: the values run from 1 to 3 and cluster at 2, so an average of 2.1 and a typical deviation of 0.7 both look right. A variance bigger than the square of the whole range means an arithmetic slip.

Worked example 2: a small binomial

A fair coin is flipped 4 times. Let $X$ be the number of heads. Find $P (X = 2)$ , $E (X)$ and $Var (X)$ .
$X \sim B (4, 0.5)$ . Exactly 2 heads: choose which 2 of the 4 flips are heads, then multiply the probabilities of that sequence: $P (X = 2) = (2 4) (0.5)^{2} (0.5)^{2} = 6 \cdot \frac{1}{16} = \frac{6}{16} = 0.375$ .
Mean: $n p = 4 \cdot 0.5 = 2$ . Variance: $n p (1 - p) = 4 \cdot 0.5 \cdot 0.5 = 1$ .
The binomial coefficient is the part students drop. Without it the answer is $\frac{1}{16}$ , the probability of one specific sequence such as HHTT, not of "two heads in any order".

The full lesson builds a table from a counting experiment, works the transformation rules, a fair-price decision, a table with an unknown, binomial "at least" and "at most" questions, and a game where you solve for the fair stake. It ends with the edge cases OMPT-E likes and the slips that cost the most marks.

How each OMPT variant tests random variables

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Not tested
OMPT-A has no statistics chapters. Nothing about expected value or distributions appears on it.
OMPT-B
64 questions
Not tested
OMPT-B extends OMPT-A into trigonometry and integration, not into probability.
OMPT-C
30 questions
Not tested
OMPT-C is numbers, algebra, linear equations and geometry. No random variables.
OMPT-D
21 questions
Not tested
OMPT-D is the calculus paper; none of its 11 chapters is statistical.
OMPT-E
40 questions
Testedroughly 5 to 7 of 40 questions across the "Random variables" and "Probability distributions" chapters
Official chapter: Random variables, 10% of the paper.
A distribution table to complete or summarise (find the missing probability, then E(X), Var(X) or the standard deviation), a rescaling question (Y = aX + b), a fair-game or expected-payout decision, and binomial items of the form "exactly k" or "at least one" with a rounded decimal answer.
OMPT-F
21 questions
Not tested
OMPT-F is business calculus without any statistics chapter. Skip this topic.
OMPT-G
26 questions
Not tested
OMPT-G covers sequences, trig, geometry, mechanics and chemistry instead of probability.

OMPT random variables practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

A random variable $X$ takes the values 1, 2, 3 with $P (X = 1) = 0.2$ , $P (X = 2) = 0.5$ , $P (X = 3) = 0.3$ . Compute $E [X]$ .

Show the worked solution

Answer: 2.1

The expectation is the probability-weighted sum of the values: $1 \cdot 0.2 + 2 \cdot 0.5 + 3 \cdot 0.3 = 0.2 + 1.0 + 0.9 = 2.1$ .

Problem 2Multiple choice

Let $Y$ be binomially distributed with $n = 10$ trials and success probability $p = 0.3$ . What is $E [Y]$ ?

A $0.3$
B $3$
C $2.1$
D $7$

Show the worked solution

Answer: B — $3$

For a binomial distribution, $E [Y] = n p = 10 \cdot 0.3 = 3$ . No sums needed — the formula does the work. (The value $2.1$ is the variance, $n p (1 - p)$ , a classic decoy.)

Problem 3Numeric answer

For the same $X$ as before (values 1, 2, 3 with probabilities 0.2, 0.5, 0.3), compute $Var (X)$ . Round to two decimals.

Show the worked solution

Answer: 0.49

First $E [X^{2}] = 1 \cdot 0.2 + 4 \cdot 0.5 + 9 \cdot 0.3 = 0.2 + 2.0 + 2.7 = 4.9$ . With $E [X] = 2.1$ , the variance is $E [X^{2}] - (E [X])^{2} = 4.9 - 4.41 = 0.49$ . The standard deviation is then a tidy $0.7$ .

Problem 4Multiple choice

Which table defines a valid probability distribution for a random variable with values 0, 1, 2?

A $P = 0.5, 0.6, - 0.1$
B $P = 0.3, 0.3, 0.3$
C $P = 0.25, 0.35, 0.40$
D $P = 0.7, 0.5, 0.2$

Show the worked solution

Answer: C — $P = 0.25, 0.35, 0.40$

A valid distribution needs every probability in $[0, 1]$ and a total of exactly 1. Only $0.25 + 0.35 + 0.40 = 1$ passes both checks. The first option has a negative entry, the second sums to 0.9, and the last sums to 1.4.

Problem 5Spot the error

A student computes the variance of a random variable with $E [X] = 3$ and $E [X^{2}] = 11$ . Step 1: recall the shortcut formula for variance in terms of $E [X^{2}]$ and $E [X]$ . Step 2: write it as $Var (X) = E [X^{2}] - E [X]$ . Step 3: substitute to get $11 - 3 = 8$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The shortcut is $Var (X) = E [X^{2}] - (E [X])^{2}$ ; Step 2 forgot to square the mean. Correctly: $11 - 9 = 2$ . Step 3 substitutes into the flawed formula without further error. Units are a good tripwire here: variance carries squared units, so both terms must be squared-scale quantities.

Problem 6Numeric answer

A basketball player makes each free throw with probability $0.4$ , independently. In 5 attempts, what is the probability of exactly 2 makes? Round to two decimals.

Show the worked solution

Answer: 0.35

This is binomial: $P (X = 2) = (2 5) (0.4)^{2} (0.6)^{3} = 10 \cdot 0.16 \cdot 0.216 = 0.3456 \approx 0.35$ .

Problem 7Numeric answer

A random variable $X$ takes the values 0, 1, 2 and 3 with probabilities 0.1, 0.4, $p$ and 0.2 respectively. Find $E (X)$ .

Show the worked solution

Answer: 1.6

The probabilities must sum to 1, so $0.1 + 0.4 + p + 0.2 = 1$ gives $p = 0.3$ . Then $E (X) = 0 (0.1) + 1 (0.4) + 2 (0.3) + 3 (0.2) = 0 + 0.4 + 0.6 + 0.6 = 1.6$ . A plausibility check: the values run from 0 to 3 with most of the weight on 1 and 2, so a mean a little above 1.5 is right.

Problem 8Multiple choice

A random variable $X$ has $Var (X) = 2$ . Let $Y = 3 X - 2$ . What is $Var (Y)$ ?

A $18$
B $6$
C $16$
D $4$

Show the worked solution

Answer: A — $18$

For $Y = a X + b$ the variance is $a^{2} Var (X)$ : the shift $b$ does nothing to the spread and the stretch $a$ is squared. So $Var (Y) = 3^{2} \cdot 2 = 18$ . In standard deviations the same fact reads $σ_{Y} = 3 σ_{X} = 32$ , and $(32)^{2} = 18$ .

Problem 9Numeric answer

A multiple-choice test has 6 questions, each with 4 options and exactly one correct answer. A student guesses every question independently. What is the probability of getting at least one question right? Round to three decimals.

Show the worked solution

Answer: 0.822

The number of correct guesses is binomial with $n = 6$ and $p = 0.25$ . "At least one" is the complement of "none": $P (X = 0) = (0.75)^{6} \approx 0.178$ , so $P (X \geq 1) = 1 - 0.178 = 0.822$ . For scale, the expected number of correct guesses is $n p = 1.5$ , which fits comfortably with an 82% chance of at least one.

Problem 10Numeric answer

A game costs $€ c$ to play. You roll two fair dice: a sum of 7 wins $€ 12$ , a sum of 11 wins $€ 9$ , and any other sum wins nothing. For which value of $c$ is the game fair?

Show the worked solution

Answer: 2.5

Fair means the expected net result is zero, so $c$ must equal the expected winnings. Over 36 ordered outcomes, a sum of 7 happens 6 ways and a sum of 11 happens 2 ways. Expected winnings: $12 \cdot \frac{6}{36} + 9 \cdot \frac{2}{36} = \frac{72 + 18}{36} = \frac{90}{36} = 2.5$ . So the fair stake is $€ 2.50$ . Any higher price makes the game a loser for the player on average.

OMPT random variables: questions students ask

Are random variables and the binomial distribution on the OMPT-E?

Yes. The OMPT-E syllabus lists "Random variables" and "Probability distributions" as two of its twelve chapters, which is why this topic carries more weight than most on that paper. Expect distribution tables, expected value and variance, linear transformations, and binomial probabilities. No other OMPT variant has these chapters.

How do I calculate expected value and variance for the OMPT?

Expected value is each value times its probability, added up. For the variance, add a column of squared values, weight those by the same probabilities to get E(X squared), then subtract the square of the expected value. The standard deviation is the square root of the variance. Write the three results as separate labelled lines; the most common error is squaring E(X) and calling it E(X squared).

What happens to the mean and variance when I rescale a random variable on the OMPT?

For Y = aX + b, the mean follows the whole transformation: E(Y) = a times E(X) plus b. The variance ignores the shift and squares the stretch: Var(Y) = a squared times Var(X). The standard deviation is multiplied by the absolute value of a. Adding a constant moves every outcome the same distance, so it changes the average and not the spread.

How do I do binomial probability questions on the OMPT without a calculator?

The numbers are chosen so that powers of the success probability stay manageable, usually p equal to 0.5, 0.2 or a simple fraction. Write P(X = k) as the binomial coefficient times p to the k times (1 minus p) to the n minus k, cancel what you can, and round only at the end. For "at least one", compute the probability of no successes and subtract it from 1.

What is the difference between expected winnings and expected net result on the OMPT?

Expected winnings is the average payout before the stake is counted. Expected net result subtracts the cost of playing. A game that pays an expected 1 euro and costs 2 euros has expected winnings of 1 euro and an expected net result of minus 1 euro. A game is fair when the expected net result is zero, and OMPT-E questions say which of the two quantities they want.

Test hubs that cover random variables:OMPT-E All 28 topics

OMPT random variables practice: lesson and 10 problems

OMPT random variables: the lesson

What the OMPT tests in random variables

Notation and answer format the OMPT expects

Worked example 1: a full table workout

Worked example 2: a small binomial

How each OMPT variant tests random variables

OMPT random variables practice problems

OMPT random variables: questions students ask

Related OMPT topics and test hubs