Discrete random variables — the probability function and the discrete uniform distribution

A discrete random variable $X$ takes the values 1, 2, 3, 4 with $p(1)=0.1$, $p(2)=0.2$, $p(3)=0.3$, $p(4)=0.3$. A student checks that every one of these four numbers lies between 0 and 1, and concludes this is a valid probability function. Is the student's check sufficient?

• No — checking each value lies between 0 and 1 is necessary but not sufficient; the values must also sum to exactly 1, and here $0.1+0.2+0.3+0.3=0.9 \neq 1$, so this is NOT a valid probability function

  Correct. Both conditions have to hold together: $0\le p(x)\le1$ for every value, AND $\sum p(x)=1$ over the whole domain. Here the four numbers individually look fine but the total is 0.9, not 1 — exactly the check a real WST01 examiner report recommends making explicitly rather than skipping (Oct 2021, Q4(d)).

• BYes — every individual value being between 0 and 1 is exactly what makes a probability function valid

  This is the incomplete check itself, stated as if it were sufficient. A set of numbers can each look like a perfectly reasonable individual probability and still fail to be a valid probability function together, if their total isn't 1.

• CNo — because $p(4)$ repeats the value 0.3 from $p(3)$, which isn't allowed

  Two different values of $x$ are entirely allowed to share the same probability — nothing about a probability function forbids repeated numbers. The actual problem here is the total, not a repeated value.

• DCannot be determined without knowing what $X$ represents

  Whether $p(1),\dots,p(4)$ form a valid probability function is a purely arithmetic question — it doesn't depend on the real-world meaning of X, only on whether the two defining conditions (each between 0 and 1; total exactly 1) hold.

Traps tested: Individual bounds treated as sufficient · Repeated probability value assumed invalid · Context assumed necessary for a purely arithmetic check

A discrete random variable $X$ has $p(1)=0.15$, $p(2)=0.35$, $p(3)=0.25$, $p(4)=0.25$. Find $F(3)$.

• $F(3) = p(1)+p(2)+p(3) = 0.15+0.35+0.25 = 0.75$

  Correct — $F(x_0)=P(X\le x_0)=\sum p(x)$ for $x\le x_0$ (WST01-verified-facts.md §1/§2a), so $F(3)$ sums every probability up to AND INCLUDING $x=3$, not stopping just short of it.

• B$F(3) = p(1)+p(2) = 0.5$, stopping before $x=3$ since $F$ means 'strictly less than'

  $F(x_0)$ is defined with $\le$, not $<$ — the cumulative distribution function includes the value $x_0$ itself. This is one of the formulae on this spec point explicitly NOT printed in the formula booklet (WST01-verified-facts.md §2a) — precisely because getting the $\le$ versus $<$ right has to come from memory, not a lookup.

• C$F(3)=p(3)=0.25$, treating $F$ as though it only looked at the single value 3

  $F$ is CUMULATIVE — a running total up to $x_0$, not the probability function evaluated at a single point. $p(3)$ alone answers 'what is P(X=3)?', a different question from F(3) = P(X≤3).

• D$F(3)=p(3)+p(4)=0.5$, summing from $x=3$ upward instead of from the bottom of the domain

  This computes $P(X\ge3)$, not $F(3)=P(X\le3)$ — the cumulative distribution function always accumulates from the SMALLEST value in the domain up to $x_0$, never the other direction.

Traps tested: Cdf boundary excluded · Cdf confused with probability function · Cdf summed in wrong direction

A discrete random variable $T$ is Discrete Uniform on $\{3,6,9,12,15\}$. Using symmetry, what is $E(T)$?

• $9$ — the values are evenly spaced and symmetric about their middle value, so $E(T)$ sits exactly at that midpoint: $\frac{3+15}{2}=9$

  Correct. For a discrete uniform distribution on evenly-spaced values, the mean can be read off by symmetry — the average of the smallest and largest value — without summing all five terms individually, exactly the shortcut a real WST01 question explicitly signals is available (Jan 2025, Q1(c): E(R) and E(B) 'by symmetry').

• B$3$, the smallest value, since that's where the distribution 'starts'

  The mean of a symmetric distribution sits at its CENTRE, not its starting value — 3 is only the minimum, not a typical outcome; the middle value here is 9.

• C$\frac{3+6+9+12+15}{5^2} = 1.8$, dividing by $n^2$ instead of $n$

  The mean of $n$ equally likely values is their sum divided by $n$ — dividing by $n^2$ has no basis in the definition of $E(X)=\sum x_iP(X=x_i)$ at all, whatever the value of $n$.

• DThe symmetry shortcut doesn't apply here, because 3, 6, 9, 12, 15 aren't consecutive integers

  The shortcut only needs the values to be evenly SPACED and symmetric about a midpoint — it doesn't require consecutive integers. 3, 6, 9, 12, 15 are spaced 3 apart throughout, exactly what the symmetry argument needs; a real WST01 question uses the same idea on the non-consecutive set {1,3,5,7} (Jan 2025, Q1).

Traps tested: Mean confused with minimum value · Wrong power of n used · Symmetry shortcut assumed to need consecutive integers

A discrete random variable $U$ is Discrete Uniform on $\{2,4,6\}$. Find $\text{Var}(U)$.

• $E(U)=4$ by symmetry; $E(U^2) = \frac{4+16+36}{3} = \frac{56}{3}$; $\text{Var}(U) = \frac{56}{3}-16 = \frac{8}{3}$

  Correct. The symmetry shortcut finds the mean quickly, but variance has no equivalent shortcut — every $x^2$ term has to be summed, exactly as a real WST01 question requires: 'Var(B) showing your working' (Jan 2025, Q1(d)) signals that no shortcut formula is expected to be quoted from memory here.

• B$\text{Var}(U)=0$, since every value in a discrete uniform distribution is 'the same' in the sense of being equally likely

  Equally LIKELY is not the same as equally SIZED — the three values 2, 4 and 6 are still spread out, and variance measures exactly that spread, not how the probabilities compare to each other.

• C$\text{Var}(U) = E(U) = 4$, reusing the mean as the variance

  $E(X)$ and $\text{Var}(X)$ are two different quantities computed from two different formulae — there's no general reason they'd be equal, and here they aren't.

• D$\text{Var}(U) = \frac{4+16+36}{3} = \frac{56}{3}$, using $E(U^2)$ alone with no subtraction step

  This is $E(U^2)$, not $\text{Var}(U)$ — the defining formula $\text{Var}(X)=E(X^2)-[E(X)]^2$ has a subtraction in it for a reason; dropping it leaves a different (and larger) quantity in variance's place.

Traps tested: Uniform likelihood confused with zero spread · Variance confused with mean · E x squared term used as variance

Two fair four-sided spinners are used independently: spinner $P$ shows 1, 2, 3, 4 and spinner $Q$ shows 2, 4, 6, 8, each value equally likely (both are discrete uniform distributions). What is $P(P=2 \text{ and } Q=2)$?

• $\frac14 \times \frac14 = \frac{1}{16}$, multiplying the two independent probabilities together

  Correct. Since the two spinners are spun independently, $P(P=2 \text{ and } Q=2) = P(P=2)\times P(Q=2) = \frac14\times\frac14 = \frac{1}{16}$ — the same combination rule a real WST01 question relies on to build the full 16-outcome table for two independent dice, one Discrete Uniform on {1,2,3,4} and the other on {1,3,5,7} (Jan 2025, Q1).

• B$\frac14+\frac14 = \frac12$, adding the two probabilities

  Adding is the rule for 'P=2 OR Q=2' (mutually exclusive events), not 'P=2 AND Q=2' on two independent spinners — for two things to both happen together, the probabilities multiply.

• C$\frac14$, since both spinners individually give the same probability of showing their respective '2'

  This only accounts for ONE spinner, not both happening together — the combined event 'P shows 2 AND Q shows 2' is rarer than either single event on its own, and the correct probability has to reflect that.

• DCannot be found without knowing whether the two spinners are correlated

  The question states the spinners are used INDEPENDENTLY — exactly the condition needed to multiply their probabilities directly, with nothing further to establish.

Traps tested: And probability computed by adding · Joint event treated as single event · Independence assumed insufficient