Discrete Random Variables

The probability distribution for the discrete random variable $X$ is given by $\Pr(X=0)=\dfrac{4}{k}$, $\Pr(X=1)=\dfrac{2k}{75}$, $\Pr(X=2)=\dfrac{k}{75}$, $\Pr(X=3)=\dfrac{2}{k}$, where $k$ is a positive real number. Show that $k = 10$ or $k = 15$.

1. 1

   All probabilities in a distribution must add to exactly $1$. Set the sum equal to $1$.

   $$\frac{4}{k} + \frac{2k}{75} + \frac{k}{75} + \frac{2}{k} = 1$$
2. 2

   Combine the like terms: the two over $k$ give $\tfrac{6}{k}$, and the two over $75$ give $\tfrac{3k}{75}=\tfrac{k}{25}$. Then multiply through by $25k$.

   $$\frac{6}{k} + \frac{k}{25} = 1 \implies 150 + k^2 = 25k$$
3. 3

   Rearrange into a quadratic and factorise. Show each line, since this is a 'show that' question.

   $$k^2 - 25k + 150 = 0 \implies (k-10)(k-15) = 0 \implies k = 10 \text{ or } k = 15$$

VCAA 2025 Mathematical Methods Exam 1, Q4a

Of travellers flying from an airport, $10\%$ check in no luggage, $40\%$ check in exactly one piece, and $50\%$ check in exactly two pieces. The probability a piece is labelled heavy is $0.234$, with masses independent. Let $W$ be the number of pieces labelled heavy checked in by a traveller. Show that $\Pr(W = 2) = 0.027$ (to three decimal places), then complete the distribution of $W$.

1. 1

   Only travellers who check in two pieces can have $W = 2$, and both pieces must be heavy. That probability is $0.5 \times 0.234^2$.

   $$\Pr(W=2) = 0.5 \times 0.234^2 = 0.02738\ldots \approx 0.027$$
2. 2

   $W = 0$ means no heavy pieces. Combine the no-luggage, one-piece-not-heavy, and two-pieces-both-light cases, using $1 - 0.234 = 0.766$.

   $$\Pr(W=0) = 0.1 + 0.4(0.766) + 0.5(0.766)^2 = 0.700$$
3. 3

   $W = 1$ means exactly one heavy piece. The remaining probability is found here, or by $1 - 0.700 - 0.027$.

   $$\Pr(W=1) = 0.4(0.234) + 0.5\cdot 2(0.234)(0.766) = 0.273$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q4c

The discrete random variable $X$ has the distribution below, where $p$ is unknown. Find $p$ and then $E(X)$. $\;x: 0,1,2,3$ with $\Pr(X=x): 0.1, 0.3, p, 0.2$.

1. 1

   All probabilities in a distribution must add to $1$. Add the known ones and subtract from $1$.

   $$p = 1 - (0.1 + 0.3 + 0.2) = 0.4$$
2. 2

   The mean is each value times its probability, all added up: $E(X) = \sum x\,\Pr(X=x)$.

   $$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2)$$
3. 3

   Work the arithmetic carefully.

   $$E(X) = 0 + 0.3 + 0.8 + 0.6 = 1.7$$

A discrete random variable $X$ takes values $1, 2, 3, 4$ with probabilities $0.2, 0.3, 0.4, 0.1$. Find $E(X)$, then the variance and the standard deviation of $X$.

1. 1

   First the mean, $E(X) = \sum x\,\Pr(X=x)$.

   $$E(X) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 2.4$$
2. 2

   Now $E(X^2)$, where you square each value but keep the same probabilities.

   $$E(X^2) = 1(0.2) + 4(0.3) + 9(0.4) + 16(0.1) = 6.6$$
3. 3

   Variance is the average of the squares minus the square of the average.

   $$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 6.6 - (2.4)^2 = 0.84$$
4. 4

   The standard deviation is the square root of the variance. Do not stop at the variance.

   $$sd(X) = \sqrt{\operatorname{Var}(X)} = \sqrt{0.84} \approx 0.917$$

A spinner lands on $1$ with probability $\tfrac{1}{2}$, on $2$ with probability $\tfrac{1}{3}$ and on $5$ with probability $\tfrac{1}{6}$. Let $X$ be the number it lands on. Find $E(X)$ and $\operatorname{Var}(X)$ exactly.

1. 1

   Check the probabilities sum to $1$: $\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6} = 1$. Then find the mean.

   $$E(X) = 1\left(\tfrac{1}{2}\right) + 2\left(\tfrac{1}{3}\right) + 5\left(\tfrac{1}{6}\right) = \tfrac{1}{2} + \tfrac{2}{3} + \tfrac{5}{6} = 2$$
2. 2

   Find $E(X^2)$ by squaring the values, keeping the probabilities.

   $$E(X^2) = 1\left(\tfrac{1}{2}\right) + 4\left(\tfrac{1}{3}\right) + 25\left(\tfrac{1}{6}\right) = \tfrac{1}{2} + \tfrac{4}{3} + \tfrac{25}{6} = 6$$
3. 3

   Apply the variance formula.

   $$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 6 - 2^2 = 2$$