Probability Distributions

Discrete Random Variables

Definition

A discrete random variable $X$ is a function that assigns a numerical value to each outcome in a countable sample space. The set of possible values is finite or countably infinite: for example, $\{0, 1, 2, \ldots, n\}$ or $\{0, 1, 2, \ldots\}$ .

Probability Mass Function (PMF)

The probability mass function of $X$ is $p(x) = P(X = x)$ Assigning a probability to each possible value. It must satisfy:

$p(x) \ge 0$ for all $x$
$\displaystyle\sum_{\mathrm{all } x} p(x) = 1$

Cumulative Distribution Function (CDF)

F(x) = P(X \le x) = \sum_{t \le x} p(t)

The CDF is non-decreasing and right-continuous, with $F(-\infty) = 0$ and $F(\infty) = 1$ . For a discrete variable it is a step function with jumps at each value in the range of $X$ . The size of each jump at $x = a$ equals $P(X = a)$ .

Expected Value

The expected value (mean) of $X$ is the probability-weighted average of all possible values:

E(X) = \mu = \sum_{\mathrm{all } x} x \cdot p(x)

This represents the long-run average if the experiment is repeated many times. For a function $g(X)$ :

E(g(X)) = \sum_{\mathrm{all } x} g(x) \cdot p(x)

A critical special case is $E(X^2) = \sum x^2 p(x)$ .

Variance and Standard Deviation

\mathrm{Var}(X) = \sigma^2 = E\!\left[(X - \mu)^2\right] = \sum_{\mathrm{all } x} (x - \mu)^2 \cdot p(x)

The computational formula is almost always more convenient:

\mathrm{Var}(X) = E(X^2) - [E(X)]^2

The standard deviation is $\sigma = \sqrt{\mathrm{Var}(X)}$ . It has the same units as $X$ and measures the typical distance of values from the mean.

Properties of Expectation and Variance

For any constant $a$ and random variable $X$ :

E(a) = a, \quad E(aX) = aE(X), \quad E(X + a) = E(X) + a

\mathrm{Var}(a) = 0, \quad \mathrm{Var}(aX) = a^2 \mathrm{Var}(X), \quad \mathrm{Var}(X + a) = \mathrm{Var}(X)

Adding a constant shifts the distribution but does not change its spread. Multiplying by $a$ scales the spread by $|a|$ .

$P(X = x) = kx$ for $x = 1, 2, 3, 4$ . Find $k$ and $E(X)$ .

$k(1 + 2 + 3 + 4) = 1 \implies 10k = 1 \implies k = 0.1$

$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$

Worked Example: E(X) and Var(X) from a Table

A random variable $X$ has the following PMF:

$x$	1	2	3	4	5
$P(X = x)$	0.1	0.2	0.3	0.25	0.15

E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.25) + 5(0.15) = 0.1 + 0.4 + 0.9 + 1.0 + 0.75 = 3.15

E(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.25) + 25(0.15) = 0.1 + 0.8 + 2.7 + 4.0 + 3.75 = 11.35

\mathrm{Var}(X) = 11.35 - 3.15^2 = 11.35 - 9.9225 = 1.4275

\sigma = \sqrt{1.4275} \approx 1.195

Binomial Distribution

Conditions

A random variable $X$ follows a binomial distribution, $X \sim B(n, p)$ When all four conditions Hold:

Fixed number of trials: exactly $n$ identical trials.
Independent trials: each trial”s outcome does not affect any other.
Two outcomes: each trial yields success (probability $p$ ) or failure (probability $q = 1-p$ ).
Constant probability: $p$ is the same for every trial.

$X$ counts the number of successes in $n$ trials.

Probability Mass Function

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n

Where $\dbinom{n}{x} = \dfrac{n!}{x!(n-x)!}$ counts the arrangements of $x$ successes among $n$ Trials.

Mean and Variance

E(X) = np, \quad \mathrm{Var}(X) = np(1-p), \quad \sigma = \sqrt{np(1-p)}

Derivation of $E(X) = np$ and $\mathrm{Var}(X) = np(1-p)$

Let $X_1, \ldots, X_n$ be indicator variables: $X_i = 1$ if trial $i$ succeeds, $X_i = 0$ otherwise. Then $X = X_1 + \cdots + X_n$ .

$E(X_i) = 1 \cdot p + 0 \cdot (1-p) = p$ So $E(X) = np$ by linearity of expectation.

$\mathrm{Var}(X_i) = E(X_i^2) - [E(X_i)]^2 = p - p^2 = p(1-p)$ So $\mathrm{Var}(X) = np(1-p)$ by Independence.

Shape

$p = 0.5$ : symmetric about $np$ .
$p \lt 0.5$ : positively skewed (right tail longer).
$p \gt 0.5$ : negatively skewed (left tail longer).

As $n$ increases the distribution approaches a bell shape (by the Central Limit Theorem). The mode of $B(n, p)$ is at $\lfloor (n+1)p \rfloor$ .

Cumulative Probabilities

On a GDC, $P(X \le k)$ is computed directly. For “at least” problems, use the complement:

P(X \ge k) = 1 - P(X \le k - 1)

Normal Approximation to the Binomial

When $n$ is large and $p$ is not too close to 0 or 1 (rule of thumb: $np \ge 5$ and $n(1-p) \ge 5$ ), The binomial can be approximated by the normal with matching mean and variance:

B(n, p) \approx N(np, np(1-p))

A continuity correction is required. For example:

P(X \le k) \approx P\!\left(Z \le \frac{k + 0.5 - np}{\sqrt{np(1-p)}}\right)

P(X = k) \approx P\!\left(\frac{k - 0.5 - np}{\sqrt{np(1-p)}} \lt Z \lt \frac{k + 0.5 - np}{\sqrt{np(1-p)}}\right)

A factory produces bulbs with 3% defect rate. $X \sim B(20, 0.03)$ is the number of defects in a sample of 20.

$P(X = 2) = \binom{20}{2}(0.03)^2(0.97)^{18} = 190 \times 0.0009 \times 0.5781 \approx 0.0988$

$P(X \le 1) = (0.97)^{20} + 20(0.03)(0.97)^{19} \approx 0.5438 + 0.3364 \approx 0.8802$

$P(X \ge 3) = 1 - P(X \le 2) \approx 1 - 0.8802 - 0.0988 = 0.0210$

$E(X) = 20(0.03) = 0.6$ , $\sigma = \sqrt{20(0.03)(0.97)} = \sqrt{0.582} \approx 0.763$

A multiple choice test has 15 questions with 5 options each. A student guesses all answers.

$X \sim B(15, 0.2)$ .

$P(X = 4) = \binom{15}{4}(0.2)^4(0.8)^{11} \approx 0.1876$

$P(X \ge 8) = 1 - P(X \le 7) \approx 0.0042$

To set a pass mark so that guessing gives at most 1% chance of passing:

$P(X \ge 7) \approx 0.0181$ and $P(X \ge 8) \approx 0.0042$ So the minimum pass mark is 8 correct.

Worked Example: Binomial Probability with Normal Approximation

A company manufactures light bulbs. On average, 8% are defective. A random sample of 100 bulbs is selected. Find the probability that more than 12 are defective.

Let $X \sim B(100, 0.08)$ .

Check conditions for normal approximation: $np = 8 \ge 5$ and $n(1-p) = 92 \ge 5$ .

\mu = 100(0.08) = 8, \quad \sigma^2 = 100(0.08)(0.92) = 7.36, \quad \sigma = 2.713

With continuity correction:

P(X \gt 12) = P(X \ge 13) \approx P\!\left(Z \gt \frac{12.5 - 8}{2.713}\right) = P(Z \gt 1.659)

\approx 1 - \Phi(1.659) \approx 1 - 0.9515 = 0.0485

There is approximately a 4.85% chance that more than 12 bulbs are defective.

Poisson Distribution

Conditions

$X \sim \mathrm{Po}(\lambda)$ models the number of events in a fixed interval of time or space when:

Events occur singly: no simultaneous events.
Independence: events in non-overlapping intervals are independent.
Constant rate: events occur at average rate $\lambda$ per unit interval.
Randomness: the count is proportional to the interval size.

Probability Mass Function

P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \ldots

Where $\lambda \gt 0$ is the mean number of events, and $e \approx 2.71828$ .

Mean and Variance

E(X) = \lambda, \quad \mathrm{Var}(X) = \lambda

That $E(X) = \mathrm{Var}(X)$ is a distinguishing feature. If observed data has mean approximately Equal to variance, a Poisson model may be appropriate.

Derivation of $E(X) = \lambda$ and $\mathrm{Var}(X) = \lambda$

$E(X) = \displaystyle\sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!} = e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^x}{(x-1)!}$

Substituting $k = x-1$ : $= e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k+1}}{k!} = \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda$ .

For variance, use $x^2 = x(x-1) + x$ : $E(X^2) = E[X(X-1)] + E(X) = \lambda^2 + \lambda$ So $\mathrm{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda$ .

Poisson as a Limit of the Binomial

If $n \to \infty$ , $p \to 0$ While $np = \lambda$ stays constant, then $B(n, p) \to \mathrm{Po}(\lambda)$ . The Poisson approximates the binomial when $n$ is large, $p$ is Small, and $np$ is moderate ( $n \ge 50$ , $p \le 0.1$ ).

Additivity

If $X \sim \mathrm{Po}(\lambda_1)$ and $Y \sim \mathrm{Po}(\lambda_2)$ are independent, then:

X + Y \sim \mathrm{Po}(\lambda_1 + \lambda_2)

If the rate is $\lambda$ per unit interval, then over $t$ intervals the count is $\mathrm{Po}(t\lambda)$ .

A typesetter makes errors at a rate of 1 per 500 characters. In a passage of 2000 characters, find the probability of at most 2 errors.

Exact: $X \sim B(2000, 1/500)$ With $\lambda = 2000/500 = 4$ .

Approximate: $X \approx \mathrm{Po}(4)$ .

$P(X \le 2) = e^{-4}\!\left(1 + 4 + \dfrac{16}{2}\right) = 13e^{-4} \approx 0.2381$

Using exact binomial: $P(X \le 2) = (499/500)^{2000} + 2000(1/500)(499/500)^{1999} + \binom{2000}{2}(1/500)^2(499/500)^{1998}$

This is computationally intensive but gives a result extremely close to 0.2381.

Worked Example: Poisson Distribution

A call centre receives calls at a rate of $\lambda = 4.2$ per 10-minute interval.

(a) Find the probability of receiving exactly 6 calls in a 10-minute interval.

P(X = 6) = \frac{e^{-4.2} \cdot 4.2^6}{6!} = \frac{e^{-4.2} \times 5489.0}{720}

= 7.624 \times e^{-4.2} \approx 7.624 \times 0.0150 \approx 0.1142

(b) Find the probability of receiving at most 3 calls.

P(X \le 3) = e^{-4.2}\!\left(1 + 4.2 + \frac{17.64}{2} + \frac{74.088}{6}\right) = e^{-4.2}(1 + 4.2 + 8.82 + 12.348)

= 26.368 \times e^{-4.2} \approx 0.3954

(c) Over a full hour (six intervals), find the probability of more than 30 calls.

Over one hour: $Y \sim \mathrm{Po}(6 \times 4.2) = \mathrm{Po}(25.2)$ .

Using the normal approximation (since $\lambda$ is large):

\mu = 25.2, \quad \sigma = \sqrt{25.2} = 5.020

P(Y \gt 30) \approx P\!\left(Z \gt \frac{30.5 - 25.2}{5.020}\right) = P(Z \gt 1.056) \approx 0.1455

Normal Distribution

Definition and Properties

$X \sim N(\mu, \sigma^2)$ has probability density function:

F(x) = \frac{1}{\sigma\sqrt{2\pi}} \, e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty \lt x \lt \infty

Key properties: bell-shaped, symmetric about $x = \mu$ Asymptotic to the $x$ -axis, total area = 1, Inflection points at $x = \mu \pm \sigma$ . The mean, median, and mode all equal $\mu$ .

$E(X) = \mu$ and $\mathrm{Var}(X) = \sigma^2$ .

For any normal variable, $P(X = a) = 0$ for any specific value $a$ (continuous distribution).

The Empirical Rule (68-95-99.7)

P(\mu - \sigma \lt X \lt \mu + \sigma) \approx 68.27\%

P(\mu - 2\sigma \lt X \lt \mu + 2\sigma) \approx 95.45\%

P(\mu - 3\sigma \lt X \lt \mu + 3\sigma) \approx 99.73\%

Standard Normal Distribution

The standard normal is $Z \sim N(0, 1)$ . Any normal variable standardises via:

Z = \frac{X - \mu}{\sigma}

The CDF is $\Phi(z) = P(Z \le z)$ . Key properties:

\Phi(-z) = 1 - \Phi(z), \quad P(Z \gt z) = 1 - \Phi(z), \quad P(-z \lt Z \lt z) = 2\Phi(z) - 1

Probability Calculations

For $X \sim N(\mu, \sigma^2)$ To find $P(a \lt X \lt b)$ Convert to $z$ -scores:

P(a \lt X \lt b) = \Phi\!\left(\frac{b - \mu}{\sigma}\right) - \Phi\!\left(\frac{a - \mu}{\sigma}\right)

On the GDC these are computed directly without manual standardisation.

Inverse Normal

Given probability $p$ The inverse normal finds $x$ such that $P(X \le x) = p$ . For the standard Normal, $z = \Phi^{-1}(p)$ . For a general normal: $x = \mu + z\sigma$ .

Finding Unknown Parameters

When $\mu$ or $\sigma$ is unknown, use standardisation with a known probability to set up simultaneous equations. Each known probability gives one equation in two unknowns; two probabilities are needed.

$X \sim B(80, 0.4)$ . Find $P(X \le 30)$ using a normal approximation.

$\mu = 80(0.4) = 32$$\sigma^2 = 80(0.4)(0.6) = 19.2$$\sigma = 4.382$ .

With continuity correction: $P(X \le 30) \approx P\!\left(Z \le \dfrac{30.5 - 32}{4.382}\right) = P(Z \le -0.342)$ .

$\approx 0.3665$

Exact binomial: $P(X \le 30) \approx 0.3642$ . The approximation is very close.

Worked Example: Normal Distribution with Unknown Parameters

Heights of a population are normally distributed. The 90th percentile is $182\,\mathrm{cm}$ and the 30th percentile is $164\,\mathrm{cm}$ . Find the mean and standard deviation.

P(X \le 182) = 0.90 \implies \frac{182 - \mu}{\sigma} = 1.282

P(X \le 164) = 0.30 \implies \frac{164 - \mu}{\sigma} = -0.524

Subtracting the second equation from the first:

\frac{18}{\sigma} = 1.806 \implies \sigma = \frac{18}{1.806} = 9.97\,\mathrm{cm}

From the first equation: $\mu = 182 - 1.282(9.97) = 182 - 12.78 = 169.2\,\mathrm{cm}$ .

So $\mu \approx 169\,\mathrm{cm}$ and $\sigma \approx 10\,\mathrm{cm}$ .

Continuous Uniform Distribution (AHL)

Definition

$X \sim U(a, b)$ has PDF:

F(x) = \frac{1}{b - a}, \quad a \le x \le b

And $f(x) = 0$ otherwise. The PDF is constant over $[a, b]$ Meaning all values in the interval are equally likely.

Mean and Variance

E(X) = \frac{a + b}{2}, \quad \mathrm{Var}(X) = \frac{(b - a)^2}{12}, \quad \sigma = \frac{b - a}{2\sqrt{3}}

Derivation

$E(X) = \displaystyle\int_a^b \frac{x}{b-a}\,dx = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2}$

$E(X^2) = \displaystyle\int_a^b \frac{x^2}{b-a}\,dx = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3}$

$\mathrm{Var}(X) = \dfrac{a^2+ab+b^2}{3} - \dfrac{(a+b)^2}{4} = \dfrac{4(a^2+ab+b^2) - 3(a^2+2ab+b^2)}{12} = \dfrac{(b-a)^2}{12}$

CDF

F(x) = \begin`\{cases}` 0 & x \lt a \\ \dfrac{x - a}{b - a} & a \le x \le b \\ 1 & x \gt b \end`\{cases}`

For any $[c, d] \subseteq [a, b]$ : $P(c \le X \le d) = \dfrac{d - c}{b - a}$ .

A bus arrives every 15 minutes. $X \sim U(0, 15)$ is the waiting time.

$P(X \gt 10) = 5/15 = 1/3$

$E(X) = 7.5$ minutes, $\sigma = \dfrac{15}{2\sqrt{3}} = \dfrac{5\sqrt{3}}{2} \approx 4.33$ minutes.

Given 5 minutes already waited, the remaining wait is $U(0, 10)$ : $P(\mathrm{wait} \ge 8) = 2/10 = 1/5$ .

Geometric Distribution (AHL)

Definition

$X \sim \mathrm{Geo}(p)$ models the number of trials needed for the first success in independent Bernoulli trials with success probability $p$ .

Probability Mass Function

P(X = x) = (1-p)^{x-1} p, \quad x = 1, 2, 3, \ldots

The first $x-1$ trials must be failures, and trial $x$ must succeed. This is the probability of exactly $x-1$ consecutive failures followed by one success.

Mean and Variance

E(X) = \frac{1}{p}, \quad \mathrm{Var}(X) = \frac{1-p}{p^2}

Derivation of $E(X) = 1/p$ and $\mathrm{Var}(X) = (1-p)/p^2$

$E(X) = p\displaystyle\sum_{x=1}^{\infty} x(1-p)^{x-1}$

Using $\displaystyle\sum_{x=1}^{\infty} xr^{x-1} = \frac{1}{(1-r)^2}$ for $|r| \lt 1$ With $r = 1-p$ :

$E(X) = p \cdot \dfrac{1}{p^2} = \dfrac{1}{p}$

For variance: $E(X^2) = E[X(X-1)] + E(X) = \dfrac{2(1-p)}{p^2} + \dfrac{1}{p} = \dfrac{2-p}{p^2}$

So $\mathrm{Var}(X) = \dfrac{2-p}{p^2} - \dfrac{1}{p^2} = \dfrac{1-p}{p^2}$ .

Useful shortcut

P(X \gt n) = (1-p)^n

The first $n$ trials must all be failures. Similarly $P(X \ge n) = (1-p)^{n-1}$ .

A basketball player has free-throw success rate 72%. $X \sim \mathrm{Geo}(0.72)$ .

$P(X = 3) = (0.28)^2(0.72) = 0.0784 \times 0.72 \approx 0.05645$

$P(X \gt 5) = (0.28)^5 \approx 0.00172$

$E(X) = 1/0.72 \approx 1.389$ attempts.

Worked Example: Geometric Distribution

A die is rolled repeatedly until a 6 appears.

$X \sim \mathrm{Geo}(1/6)$ .

(a) Find the probability that the first 6 appears on the 4th roll.

P(X = 4) = \left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{1296} \approx 0.0965

(b) Find the probability that at least 10 rolls are needed.

P(X \ge 10) = (1 - p)^{10-1} = \left(\frac{5}{6}\right)^9 = 0.1938

(c) Find the expected number of rolls.

E(X) = \frac{1}{p} = \frac{1}{1/6} = 6

On average, 6 rolls are needed to get the first 6.

Negative Binomial Distribution (AHL)

Definition

$X \sim \mathrm{NB}(r, p)$ models the number of trials needed to obtain exactly $r$ successes. The Geometric distribution is the special case $\mathrm{NB}(1, p)$ .

Probability Mass Function

P(X = x) = \binom{x-1}{r-1} p^r (1-p)^{x-r}, \quad x = r, r+1, r+2, \ldots

In the first $x-1$ trials there are $r-1$ successes (in $\dbinom{x-1}{r-1}$ ways), and trial $x$ is the $r$ -th success.

Mean and Variance

E(X) = \frac{r}{p}, \quad \mathrm{Var}(X) = \frac{r(1-p)}{p^2}

Note the parallel with geometric: multiplying $r$ by a factor scales both $E(X)$ and $\mathrm{Var}(X)$ by the same factor.

A coin has $P(\mathrm{heads}) = 0.4$ . $X \sim \mathrm{NB}(3, 0.4)$ counts flips for 3 heads.

$P(X = 7) = \dbinom{6}{2}(0.4)^3(0.6)^4 = 15 \times 0.064 \times 0.1296 \approx 0.1244$

$E(X) = 3/0.4 = 7.5$ , $\mathrm{Var}(X) = 3(0.6)/0.16 = 11.25$ $\sigma = \sqrt{11.25} \approx 3.354$ .

Central Limit Theorem (AHL)

Statement

If $X_1, X_2, \ldots, X_n$ are independent and identically distributed with mean $\mu$ and variance $\sigma^2$ Then for large $n$ :

\bar{X}_n \sim N\!\left(\mu, \frac{\sigma^2}{n}\right)

This holds regardless of the shape of the original distribution. The rule of thumb is $n \ge 30$ .

Distribution of the Sum

The sum $S_n = X_1 + \cdots + X_n$ is approximately $S_n \sim N(n\mu, n\sigma^2)$ for large $n$ .

Standard Error

\mathrm{SE}(\bar{X}) = \frac{\sigma}{\sqrt{n}}

As $n$ increases, the standard error decreases: larger samples give more precise estimates of the population mean.

$X \sim U(2, 10)$ . Sample of 50 observations. Find $P(\mathrm{sum} \gt 310)$ .

$\mu = 6$ , $\sigma^2 = 64/12 = 16/3$ . Sum has mean $300$ and variance $50(16/3) = 800/3$ .

$P\!\left(Z \gt \dfrac{10}{\sqrt{800/3}}\right) = P(Z \gt 0.612) \approx 0.2704$ .

Confidence Intervals (AHL)

Concept

A $C\%$ confidence interval gives a range of plausible values for an unknown population parameter. If the sampling process were repeated many times, approximately $C\%$ of constructed intervals would contain the true parameter. The confidence level does not mean there is a $C\%$ probability that the parameter lies in any particular interval.

Confidence Interval for the Mean ( $\sigma$ known)

\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

Where $z_{\alpha/2}$ satisfies $P(Z \gt z_{\alpha/2}) = \alpha/2$ and $\alpha = 1 - C/100$ .

Confidence Level	$z_{\alpha/2}$
90%	1.645
95%	1.960
99%	2.576

When $\sigma$ is unknown and $n$ is large ( $n \ge 30$ ), replace $\sigma$ with the sample standard Deviation $s$ .

Margin of Error and Sample Size

Margin of error: $E = z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}$ . To halve $E$ Quadruple $n$ .

Required sample size for margin $E$ : $n = \left(\dfrac{z_{\alpha/2} \cdot \sigma}{E}\right)^2$ (round up to the next integer).

Bottle volumes: $N(\mu, 25)$ , $\sigma = 5$ ml. Sample of 25 gives $\bar{x} = 498$ ml.

95% CI: $498 \pm 1.960 \times 5/\sqrt{25} = 498 \pm 1.96$ So $(496.04, 499.96)$ ml.

For margin 1 ml at 95%: $n = (1.960 \times 5/1)^2 = 96.04$ Round up to 97.

Combining Random Variables

Linear Combinations

For any random variables $X$$Y$ and constants $a$$b$ :

E(aX + bY) = aE(X) + bE(Y)

This is the linearity of expectation and holds always, even without independence.

Variance of Sums

For independent $X$ and $Y$ :

\mathrm{Var}(aX + bY) = a^2\mathrm{Var}(X) + b^2\mathrm{Var}(Y)

\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y), \quad \mathrm{Var}(X - Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)

Note the plus sign even for differences: subtracting a variable still adds variability.

The general formula (not necessarily independent):

\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\mathrm{Cov}(X, Y)

Where $\mathrm{Cov}(X, Y) = E(XY) - E(X)E(Y) = 0$ when $X$ and $Y$ are independent.

Linearity of expectation always holds. The simple variance formula $\mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ requires independence.

Independent Copies

If $X_1, \ldots, X_n$ are iid with mean $\mu$ and variance $\sigma^2$ :

E(X_1 + \cdots + X_n) = n\mu, \quad \mathrm{Var}(X_1 + \cdots + X_n) = n\sigma^2

E(\bar{X}) = \mu, \quad \mathrm{Var}(\bar{X}) = \frac{\sigma^2}{n}

Combining Normal Variables

If $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$ are independent, then:

AX + bY \sim N(a\mu_X + b\mu_Y, a^2\sigma_X^2 + b^2\sigma_Y^2)

This is exact (not an approximation) for normal variables, and requires no CLT.

Bus ride $X \sim N(25, 16)$ Walk $Y \sim N(10, 9)$ Independent.

$X + Y \sim N(35, 25)$ . $P(X + Y \gt 40) = P(Z \gt 1) = 0.1587$ .

Machine A produces rods: $X \sim N(50.0, 0.04)$ . Machine B: $Y \sim N(50.2, 0.09)$ .

$X - Y \sim N(-0.2, 0.13)$ . $P(X - Y \gt 0) = P\!\left(Z \gt \dfrac{0.2}{\sqrt{0.13}}\right) = P(Z \gt 0.555) \approx 0.2894$ .

Worked Example: Combining Random Variables

$X \sim B(12, 0.3)$ and $Y \sim \mathrm{Po}(5)$ are independent. Find:

(a) $E(3X - 2Y)$

E(3X - 2Y) = 3E(X) - 2E(Y) = 3(12 \times 0.3) - 2(5) = 3(3.6) - 10 = 10.8 - 10 = 0.8

(b) $\mathrm{Var}(3X - 2Y)$

\mathrm{Var}(3X - 2Y) = 9\mathrm{Var}(X) + 4\mathrm{Var}(Y) = 9(12 \times 0.3 \times 0.7) + 4(5)

= 9(2.52) + 20 = 22.68 + 20 = 42.68

Note: the variance of the difference uses addition (plus signs for both terms), and the constants are squared.

IB Exam-Style Questions

Question 1 (Paper 1)

$X \sim B(20, 0.35)$ . Find $P(5 \le X \le 8)$ .

$P(5 \le X \le 8) = P(X \le 8) - P(X \le 4) \approx 0.7625 - 0.1260 = 0.6365$

Question 2 (Paper 1)

$X \sim \mathrm{Po}(4.2)$ . Find $P(X \ge 3)$ .

$P(X \ge 3) = 1 - P(X \le 2) = 1 - e^{-4.2}(1 + 4.2 + 8.82) \approx 1 - 0.2103 = 0.7897$

Question 3 (Paper 2)

Daily rainfall: $X \sim N(2.8, 1.44)$ (mean 2.8 mm, $\sigma = 1.2$ mm).

$P(X \gt 4) = P(Z \gt 1) = 0.1587$

Expected days per year exceeding 4 mm: $365 \times 0.1587 \approx 58$ days.

Rainfall exceeded on only 5% of days: $x = 2.8 + 1.645(1.2) = 4.774$ mm.

Question 4 (Paper 2, AHL)

$X \sim U(0, a)$ has $P(X \gt 3) = 0.4$ . Find $a$ and $\mathrm{Var}(X)$ .

$\dfrac{a-3}{a} = 0.4 \implies 0.6a = 3 \implies a = 5$

$\mathrm{Var}(X) = 25/12 \approx 2.083$

Question 5 (Paper 2, AHL)

$X \sim \mathrm{Geo}(0.15)$ . Find the smallest $n$ with $P(X \le n) \ge 0.8$ .

$P(X \le n) = 1 - 0.85^n \ge 0.8 \implies 0.85^n \le 0.2$

$n \ge \ln(0.2)/\ln(0.85) \approx 9.90$ So $n = 10$ .

Question 6 (Paper 2, AHL)

Component lengths: $N(\mu, 0.25)$$\sigma = 0.5$ mm. Sample of 30 gives $\bar{x} = 100.2$ mm.

90% CI: $100.2 \pm 1.645 \times 0.5/\sqrt{30} = 100.2 \pm 0.150$ So $(100.05, 100.35)$ mm.

The claim $\mu = 100$ mm is not supported at 90% confidence, since 100 falls below the interval.

Question 7 (Paper 2, AHL)

$X \sim \mathrm{NB}(4, 0.25)$ . Find $P(X = 10)$ and $E(X)$ .

$P(X = 10) = \dbinom{9}{3}(0.25)^4(0.75)^6 = 84 \times 0.003906 \times 0.1780 \approx 0.0584$

$E(X) = 4/0.25 = 16$

Question 8 (Paper 2, AHL)

The masses of male students are $N(72, 36)$ and female students are $N(58, 25)$ Independent. Find The probability that a randomly chosen male is heavier than a randomly chosen female.

Let $M \sim N(72, 36)$ and $F \sim N(58, 25)$ . Then $D = M - F \sim N(72-58, 36+25) = N(14, 61)$ .

$P(D \gt 0) = P\!\left(Z \gt \dfrac{0 - 14}{\sqrt{61}}\right) = P(Z \gt -1.793) = \Phi(1.793) \approx 0.9636$

Summary of Distributions

Discrete Distributions

Distribution	Notation	PMF	$E(X)$	$\mathrm{Var}(X)$	Support
Binomial	$B(n, p)$	$\dbinom{n}{x}p^x(1-p)^{n-x}$	$np$	$np(1-p)$	$0, 1, \ldots, n$
Poisson	$\mathrm{Po}(\lambda)$	$\dfrac{e^{-\lambda}\lambda^x}{x!}$	$\lambda$	$\lambda$	$0, 1, 2, \ldots$
Geometric (AHL)	$\mathrm{Geo}(p)$	$(1-p)^{x-1}p$	$\dfrac{1}{p}$	$\dfrac{1-p}{p^2}$	$1, 2, 3, \ldots$
Neg. Binomial (AHL)	$\mathrm{NB}(r, p)$	$\dbinom{x-1}{r-1}p^r(1-p)^{x-r}$	$\dfrac{r}{p}$	$\dfrac{r(1-p)}{p^2}$	$r, r+1, \ldots$

Continuous Distributions

Distribution	Notation	PDF	$E(X)$	$\mathrm{Var}(X)$	Support
Normal	$N(\mu, \sigma^2)$	$\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$	$\mu$	$\sigma^2$	$(-\infty, \infty)$
Uniform (AHL)	$U(a, b)$	$\dfrac{1}{b-a}$	$\dfrac{a+b}{2}$	$\dfrac{(b-a)^2}{12}$	$[a, b]$

Key Relationships

Relationship	Condition
$B(n, p) \approx \mathrm{Po}(np)$	$n$ large, $p$ small, $np$ moderate
$B(n, p) \approx N(np, np(1-p))$	$np \ge 5$$n(1-p) \ge 5$ With continuity correction
$\mathrm{Geo}(p) = \mathrm{NB}(1, p)$	Special case
$X + Y \sim \mathrm{Po}(\lambda_1 + \lambda_2)$	Independent Poisson variables
$aX + bY \sim N(a\mu_X + b\mu_Y, a^2\sigma_X^2 + b^2\sigma_Y^2)$	Independent normal variables
$\bar{X}_n \approx N(\mu, \sigma^2/n)$	CLT, large $n$
$E(aX + bY) = aE(X) + bE(Y)$	Always
$\mathrm{Var}(aX + bY) = a^2\mathrm{Var}(X) + b^2\mathrm{Var}(Y)$	$X$$Y$ independent

Intuition

Probability distributions are recipes for uncertainty. The binomial distribution counts successes in fixed trials, flipping a coin ten times and counting heads. The Poisson distribution counts rare events in continuous time or space, the number of customers arriving at a shop each hour. The normal distribution is the bell curve that emerges logically when many small independent factors combine, like heights of adults shaped by thousands of genetic and environmental influences. The Central Limit Theorem is the deep reason the normal distribution appears everywhere: sample means tend toward normality regardless of the original distribution, which is why statistics works at all.

Common Pitfalls

Confusing $p$ and $\lambda$ : For Poisson, $\lambda$ is a rate, not a probability. Unlike binomial $p$ There is no upper bound of 1 on $\lambda$ .
Forgetting conditions: Before applying a distribution, verify all conditions. For binomial: fixed $n$ Independence, two outcomes, constant $p$ .
Variance of differences: $\mathrm{Var}(X - Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ (plus, not minus) for independent variables.
Continuity correction: When approximating a discrete distribution with a continuous one, apply a continuity correction. For example, $P(X \le 5)$ becomes $P(X \lt 5.5)$ under the normal approximation.
Standardisation direction: $\Phi(z)$ goes from $z$ -score to probability; $\Phi^{-1}(p)$ goes from probability to $z$ -score.
Geometric support: $X \sim \mathrm{Geo}(p)$ counts trials starting from 1 (IB convention).
Poisson additivity: Requires independence. If events are correlated, the sum is not Poisson.
Confidence interval interpretation: A 95% CI does not mean there is a 95% probability that $\mu$ lies in the interval. It means 95% of similarly constructed intervals contain $\mu$ .
Squaring constants in variance: $\mathrm{Var}(3X) = 9\mathrm{Var}(X)$ Not $3\mathrm{Var}(X)$ .

Problem Set

Problem 1

A discrete random variable $X$ has PMF $P(X = x) = \frac{x + 1}{15}$ for $x = 0, 1, 2, 3, 4$ . Find $E(X)$$\mathrm{Var}(X)$ And $P(X \ge 2)$ .

Solution

Verify: $\sum_{x=0}^{4}\frac{x+1}{15} = \frac{1+2+3+4+5}{15} = \frac{15}{15} = 1$ .

E(X) = 0\!\left(\frac{1}{15}\right) + 1\!\left(\frac{2}{15}\right) + 2\!\left(\frac{3}{15}\right) + 3\!\left(\frac{4}{15}\right) + 4\!\left(\frac{5}{15}\right)

= \frac{0 + 2 + 6 + 12 + 20}{15} = \frac{40}{15} = \frac{8}{3} \approx 2.667

E(X^2) = 0 + 1\!\left(\frac{2}{15}\right) + 4\!\left(\frac{3}{15}\right) + 9\!\left(\frac{4}{15}\right) + 16\!\left(\frac{5}{15}\right) = \frac{0 + 2 + 12 + 36 + 80}{15} = \frac{130}{15} = \frac{26}{3}

\mathrm{Var}(X) = \frac{26}{3} - \left(\frac{8}{3}\right)^2 = \frac{26}{3} - \frac{64}{9} = \frac{78 - 64}{9} = \frac{14}{9} \approx 1.556

P(X \ge 2) = \frac{3}{15} + \frac{4}{15} + \frac{5}{15} = \frac{12}{15} = \frac{4}{5} = 0.8

If you get this wrong, revise: Discrete Random Variables section.

Problem 2

$X \sim B(25, 0.35)$ . Find $P(X = 10)$$P(X \le 5)$ And $P(X \ge 15)$ .

Solution

P(X = 10) = \binom{25}{10}(0.35)^{10}(0.65)^{15} \approx 0.1268

P(X \le 5) = \sum_{x=0}^{5}\binom{25}{x}(0.35)^x(0.65)^{25-x} \approx 0.0334

P(X \ge 15) = 1 - P(X \le 14) \approx 1 - 0.9752 = 0.0248

If you get this wrong, revise: Binomial Distribution section.

Problem 3

A bookshop sells an average of 3.2 rare books per week. $X \sim \mathrm{Po}(3.2)$ is the number sold in a week. Find $P(X = 4)$ , $P(X = 0)$ And $P(X \gt 5)$ .

Solution

P(X = 4) = \frac{e^{-3.2} \cdot 3.2^4}{4!} = \frac{104.858 \times e^{-3.2}}{24} \approx 0.1781

P(X = 0) = e^{-3.2} \approx 0.0408

P(X \gt 5) = 1 - P(X \le 5) = 1 - e^{-3.2}\!\left(1 + 3.2 + \frac{10.24}{2} + \frac{32.768}{6} + \frac{104.858}{24} + \frac{335.544}{120}\right)

= 1 - e^{-3.2}(1 + 3.2 + 5.12 + 5.461 + 4.369 + 2.796) = 1 - e^{-3.2}(21.946) \approx 1 - 0.8955 = 0.1045

If you get this wrong, revise: Poisson Distribution section.

Problem 4

Exam scores follow $N(65, 64)$ (mean 65, variance 64). Find the probability that a randomly chosen Student scores above 75, and the score that is exceeded by only 10% of students.

Solution

$\mu = 65$ , $\sigma = \sqrt{64} = 8$ .

P(X \gt 75) = P\!\left(Z \gt \frac{75 - 65}{8}\right) = P(Z \gt 1.25) = 1 - \Phi(1.25) \approx 1 - 0.8944 = 0.1056

For the 90th percentile (exceeded by only 10%):

P(X \le x) = 0.90 \implies \frac{x - 65}{8} = 1.282 \implies x = 65 + 1.282(8) = 75.26

A score of approximately 75.3 is exceeded by only 10% of students.

If you get this wrong, revise: Normal Distribution section.

Problem 5

The waiting time for a train is uniformly distributed between 0 and 12 minutes. Find the probability that the waiting time is (a) less than 5 minutes, (b) between 7 and 10 minutes, (c) more than 8 minutes given that it has already been 3 minutes.

Solution

$X \sim U(0, 12)$ .

(a) $P(X \lt 5) = 5/12 \approx 0.4167$

(b) $P(7 \lt X \lt 10) = (10 - 7)/12 = 3/12 = 0.25$

(c) Given 3 minutes already waited, the remaining time is $U(0, 9)$ (memoryless property of the Uniform distribution):

$P(\mathrm{remaining} \gt 5) = 4/9 \approx 0.4444$

Alternatively: $P(X \gt 8 \mid X \gt 3) = P(X \gt 8)/P(X \gt 3) = (4/12)/(9/12) = 4/9$ .

If you get this wrong, revise: Continuous Uniform Distribution section.

Problem 6

$X \sim \mathrm{Geo}(0.25)$ . Find the smallest $n$ such that $P(X \le n) \ge 0.95$ .

Solution

P(X \le n) = 1 - (1 - p)^n = 1 - 0.75^n \ge 0.95

0.75^n \le 0.05

n \ln(0.75) \le \ln(0.05)

n \ge \frac{\ln(0.05)}{\ln(0.75)} = \frac{-2.996}{-0.288} = 10.40

So $n = 11$ trials are needed.

If you get this wrong, revise: Geometric Distribution section.

Problem 7

$X \sim \mathrm{NB}(3, 0.2)$ . Find $P(X = 8)$ and $\mathrm{Var}(X)$ .

Solution

P(X = 8) = \binom{7}{2}(0.2)^3(0.8)^5 = 21 \times 0.008 \times 0.32768 = 0.05505

E(X) = \frac{3}{0.2} = 15

\mathrm{Var}(X) = \frac{3(0.8)}{0.04} = \frac{2.4}{0.04} = 60

\sigma = \sqrt{60} \approx 7.75

If you get this wrong, revise: Negative Binomial Distribution section.

Problem 8

The masses of packets of sugar are normally distributed with mean $500\,\mathrm{g}$ and standard Deviation $5\,\mathrm{g}$ . A sample of 36 packets is selected. Find the probability that the sample Mean is between $498\,\mathrm{g}$ and $503\,\mathrm{g}$ .

Solution

By the CLT:

\bar{X} \sim N\!\left(500, \frac{25}{36}\right)

\sigma_{\bar{X}} = \frac{5}{6} \approx 0.833

P(498 \lt \bar{X} \lt 503) = P\!\left(\frac{498 - 500}{5/6} \lt Z \lt \frac{503 - 500}{5/6}\right) = P(-2.4 \lt Z \lt 3.6)

= \Phi(3.6) - \Phi(-2.4) = 0.9998 - 0.0082 = 0.9916

If you get this wrong, revise: Central Limit Theorem section.

Problem 9

A 95% confidence interval for the mean diameter of bolts is $(10.02\,\mathrm{mm}, 10.18\,\mathrm{mm})$ Based on a sample of size 50. The population standard deviation is known to be $\sigma = 0.4\,\mathrm{mm}$ . Find the sample mean and verify the confidence interval.

Solution

The sample mean is the midpoint of the interval:

\bar{x} = \frac{10.02 + 10.18}{2} = 10.10\,\mathrm{mm}

The margin of error is half the width:

E = \frac{10.18 - 10.02}{2} = 0.08\,\mathrm{mm}

Verify: $E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 1.960 \times \frac{0.4}{\sqrt{50}} = 1.960 \times 0.0566 = 0.1109$

The calculated margin of error ( $0.1109$ ) exceeds the stated margin ( $0.08$ ). This suggests the confidence interval was constructed with a different confidence level or the stated $\sigma$ does not match the data. If we solve for the confidence level that gives $E = 0.08$ :

z_{\alpha/2} = \frac{0.08}{0.4/\sqrt{50}} = \frac{0.08}{0.0566} = 1.413

This corresponds to approximately 84% confidence, not 95%.

If you get this wrong, revise: Confidence Intervals section.

Problem 10

$X \sim B(15, 0.4)$ and $Y \sim B(20, 0.3)$ are independent. Find $E(X + Y)$ $\mathrm{Var}(X - Y)$ And $P(X + Y = 10)$ .

Solution

E(X) = 15(0.4) = 6, \quad E(Y) = 20(0.3) = 6

E(X + Y) = 6 + 6 = 12

\mathrm{Var}(X) = 15(0.4)(0.6) = 3.6, \quad \mathrm{Var}(Y) = 20(0.3)(0.7) = 4.2

\mathrm{Var}(X - Y) = 3.6 + 4.2 = 7.8

For $P(X + Y = 10)$ Enumerate pairs $(x, y)$ where $x + y = 10$$0 \le x \le 15$$0 \le y \le 20$ :

This requires summing over $x = 0$ to $x = 10$ :

P(X + Y = 10) = \sum_{x=0}^{10} P(X = x)P(Y = 10 - x)

This is computationally intensive without a GDC, but the key principle is clear: since $X$ and $Y$ are Independent binomial variables with the same success probability ( $p = 0.3$ and $p = 0.4$ differ, So the sum is not binomial), the distribution of $X + Y$ must be found by convolution.

If you get this wrong, revise: Combining Random Variables section.

Problem 11

The lifetimes of batteries are normally distributed with mean $500\,\mathrm{hours}$ and standard Deviation $50\,\mathrm{hours}$ . Find the probability that a randomly selected battery lasts more than $550\,\mathrm{hours}$ . If four batteries are selected independently, find the probability that at Least three last more than $550\,\mathrm{hours}$ .

Solution

P(X \gt 550) = P\!\left(Z \gt \frac{550 - 500}{50}\right) = P(Z \gt 1) = 1 - 0.8413 = 0.1587

Let $Y$ be the number (out of 4) lasting more than 550 hours. $Y \sim B(4, 0.1587)$ .

P(Y \ge 3) = P(Y = 3) + P(Y = 4)

= \binom{4}{3}(0.1587)^3(0.8413) + (0.1587)^4

= 4(0.003997)(0.8413) + 0.000635 = 0.01345 + 0.000635 = 0.01409

Approximately 1.4% chance that at least three out of four batteries last more than 550 hours.

If you get this wrong, revise: Normal Distribution and Binomial Distribution sections.

Problem 12

Use the Poisson approximation to the binomial to estimate the probability of getting 3 or more Sixes when rolling a fair die 60 times.

Solution

$X \sim B(60, 1/6)$ . $\lambda = np = 60/6 = 10$ .

Approximate: $X \approx \mathrm{Po}(10)$ .

Check conditions: $n = 60 \ge 50$ , $p = 1/6 \le 0.1$ ? No, $p = 0.167 \gt 0.1$ . The Poisson Approximation is less accurate here but still usable as an estimate.

P(X \ge 3) = 1 - P(X \le 2) = 1 - e^{-10}\!\left(1 + 10 + \frac{100}{2}\right)

= 1 - 61e^{-10} = 1 - 61(0.0000454) = 1 - 0.00277 = 0.9972

Exact binomial: $P(X \le 2) = \binom{60}{0}(5/6)^{60} + \binom{60}{1}(1/6)(5/6)^{59} + \binom{60}{2}(1/6)^2(5/6)^{58}$

This gives approximately $P(X \le 2) \approx 0.00268$ So $P(X \ge 3) \approx 0.9973$ . The approximation is quite close despite $p \gt 0.1$ because $\lambda = 10$ is moderate.

If you get this wrong, revise: Poisson as a Limit of the Binomial section.

A-Level Probability: Mathematics
DSE Probability: Probability
University Probability and Statistics: Probability and Statistics

Worked Examples

Example 1: Binomial expansion

Find the first three terms in the expansion of $(1 + 2x)^{10}$ in ascending powers of $x$ .

Solution:

(1 + 2x)^{10} = \binom{10}{0}(1)^{10}(2x)^0 + \binom{10}{1}(1)^9(2x)^1 + \binom{10}{2}(1)^8(2x)^2

= 1 + 10(2x) + 45(4x^2) = 1 + 20x + 180x^2

Summary

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ ; $P(A|B) = \frac{P(A \cap B)}{P(B)}$
Mutually exclusive: $P(A \cap B) = 0$ ; independent: $P(A \cap B) = P(A)P(B)$
Binomial distribution: $X \sim B(n, p)$ ; $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$
Normal distribution: standardise with $Z = \frac{X - \mu}{\sigma}$

Cross-References

Topic	Site	Link
[Statistical Distributions]	A-Level	View
[Statistical Distributions]	IB	View

Probability Distributions

Discrete Random Variables

Definition

Probability Mass Function (PMF)

Cumulative Distribution Function (CDF)

Expected Value

Variance and Standard Deviation

Properties of Expectation and Variance

Problem Set

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Related Content at Other Levels

Worked Examples

Summary

Cross-References

See Also