OMPT drill — Probability & statistics

Random variables

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

Lesson

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 lottery ticket. Its behaviour is summarised in a distribution table: each possible value alongside the probability of seeing it, with the probabilities summing to exactly 1. That table is the whole object. Every question in this topic is answered by reading it correctly.

Two numbers compress the table further. The expected value $E (X)$ is the long-run average: each value weighted by its probability, added up. The variance measures spread around that average, and its square root, the standard deviation, puts the spread back in the same units as $X$ . The arithmetic in this topic is genuinely 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 topic’s signature error.

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