N1, N3 - Discrete and binomial distributions | AQA A-Level Mathematics

Describe separate random outcomes with a probability distribution, then recognise when a count has a binomial model. N1 develops exact and cumulative probabilities; N3 helps you judge the assumptions and interpret binomial mean and spread.

Discrete probability distributions

Can a model predict an average of fourteen delayed parcels without claiming that the next delivery contains fourteen delays? First separate the possible outcomes from the probabilities attached to them. A small collection desk might have no customers waiting, one, two or three. The number waiting is random; its possible values are separate.

You already know

Mutually exclusive outcomes cannot occur together. An exhaustive list covers every possibility, so its probabilities add to $1$ . Add probabilities when an event contains several mutually exclusive outcomes.

A random variable assigns a number to each outcome of a random process. Capital $X$ names the variable; lowercase $x$ names one possible value. Thus $X=2$ is an event, not a claim that $X$ is always $2$ .

Discrete random variable

A discrete random variable takes separate possible values. Its probability distribution specifies those values and their probabilities. A probability function is written $p(x)=\mathrm P(X=x)$ .

Counts have non-negative integer values, often written $\mathbb Z_0^+=\{0,1,2,\ldots\}$ . Discrete variables need not be counts: possible cash prizes can be separate amounts. A distribution can be given by a table or by a function with a stated range of values. A listed value may have probability zero, but negative probabilities are impossible. Check both the range of each probability and the total. These probability rules are not in the formula booklet: learn them.

Worked example 1: completing a distribution

Question. The number $X$ waiting at a desk has the distribution below. Find $k$ , $\mathrm P(X\geq2)$ and the probability that $X$ is odd.

$x$	$0$	$1$	$2$	$3$
$\mathrm P(X=x)$	$k$	$2k$	$3k$	$4k$

Plan. Use total probability to determine the scale, then add the probabilities of the included values.

\begin{aligned} k+2k+3k+4k&=1 && \text{exhaustive outcomes} \\ 10k&=1 && \text{collect terms} \\ k&=\frac1{10} && \text{divide by }10 \\ \mathrm P(X\geq2)&=3k+4k=0.7 && \text{include }2,3 \\ \mathrm P(X\text{ is odd})&=2k+4k=0.6 && \text{include }1,3 \end{aligned}

Why this step? An event selects outcomes; it is their probabilities that you add, not the outcome values.

Answer. $k=0.1$ , $\mathrm P(X\geq2)=0.7$ and the odd-value probability is $0.6$ .

Check. The probabilities $0.1,0.2,0.3,0.4$ are all valid and total $1$ .

Change only the last weight from $4k$ to $6k$ . That changes the scale of every probability, so you must find $k$ again.

Worked example 2: one weight changes

Question. A second desk has probabilities $k,2k,3k,6k$ for $X=0,1,2,3$ , respectively, and zero otherwise. Find $k$ , $\mathrm P(X\geq2)$ and the odd-value probability.

Plan. Normalise the new weights before selecting the same two events.

\begin{aligned} (1+2+3+6)k&=1 && \text{total probability} \\ k&=\frac1{12} && \text{solve} \\ \mathrm P(X\geq2)&=\frac{3+6}{12}=\frac34 && \text{include }2,3 \\ \mathrm P(X\text{ is odd})&=\frac{2+6}{12}=\frac23 && \text{include }1,3 \end{aligned}

Why this step? The weights describe relative sizes; none is a probability until the total has been made $1$ .

Answer. $k=1/12$ , $\mathrm P(X\geq2)=3/4$ and the odd-value probability is $2/3$ .

Check. $(1+2+3+6)/12=1$ , and both event probabilities lie between $0$ and $1$ .

An odd event is not necessarily a single outcome. Here it contains both $1$ and $3$ .

Watch out

Wrong: $\mathrm P(X\geq2)=2+3=5$ . These are outcome values, and a probability cannot exceed $1$ . Right: add $\mathrm P(X=2)+\mathrm P(X=3)$ .

You may see $\mathrm E(X)$ for a mean and $\operatorname{Var}(X)$ for a variance. General discrete-distribution moment calculations are not needed here; the specific binomial summaries later are required. Try another normalisation, this time with a probability function.

Binomial conditions and exact probabilities

Six devices each either pass a check or fail it. Counting passes suggests a binomial model only if the process meets four conditions: a fixed number of trials, two categories per trial, independent trials and the same success probability on every trial.

Binomial model

$X\sim \mathrm B(n,p)$ means that $X$ counts successes in $n$ independent trials, each with success probability $p$ . The symbol $\sim$ means “is distributed as”. Here $n$ is a positive integer and $0\leq p\leq1$ .

“Success” means the category you count: a faulty device can be a statistical success. Several physical outcomes can be grouped into success and failure, such as blue or not blue on a multicoloured spinner. Write $q=1-p$ for failure probability.

Independence concerns the relationship between trials; constant $p$ concerns their individual chances. Neither follows merely from calling a sample random. Devices from one faulty batch may fail together, while devices from different production lines may have different fault probabilities. Sampling a small bag without replacement usually changes conditional probabilities. Replacement and fresh independent random selection restore the same chances. Treating a tiny sample from a very large population as approximately independent requires a contextual justification. Taking shots until the third goal gives a random trial count, so it is not a fixed-trial binomial experiment.

For $0<p<1$ , one particular sequence with $r$ successes and $n-r$ failures has probability $p^r(1-p)^{n-r}$ . There are $\binom nr$ possible success positions.

Exact binomial probability

For $X\sim \mathrm B(n,p)$ ,

\mathrm P(X=r)=\binom nr p^r(1-p)^{n-r},\qquad r=0,1,\ldots,n

with $\displaystyle\binom nr={}_nC_r=\frac{n!}{r!(n-r)!}$ .
In the formula booklet.

For a positive integer $n$ , $n!=n(n-1)\cdots1$ , and $0!=1$ . The same coefficients occur in $(q+p)^n$ , whose value $1$ confirms that all binomial probabilities sum to $1$ . Recognise the boundary cases directly: $p=0$ makes $X=0$ certain; $p=1$ makes $X=n$ certain. The support is always within $0,1,\ldots,n$ .

Worked example 3: exact success count

Question. Six independent devices each pass with probability $0.3$ . Let $X$ count passes. Find $\mathrm P(X=2)$ .

Plan. State the count model, then use the exact binomial probability function.

\begin{aligned} X&\sim \mathrm B(6,0.3) && \text{fixed independent checks} \\ \mathrm P(X=2)&=\binom62(0.3)^2(0.7)^4 && \text{two passes, four failures} \\ &=15(0.3)^2(0.7)^4 && \text{choose pass positions} \\ &=0.324135 && \text{calculator evaluation} \end{aligned}

Why this step? Multiplying by $15$ includes every arrangement, rather than just two passes followed by four failures.

Answer. $\mathrm P(X=2)=0.324135$ .

Check. The calculator's exact binomial function with $n=6,p=0.3,x=2$ gives the same result.

Change only the requested count from two passes to three. Both the arrangement count and the success/failure powers must change.

Worked example 4: a different exact count

Question. For the same six independent devices, each passing with probability $0.3$ , find $\mathrm P(X=3)$ .

Plan. Keep the model and change the event entered into the exact probability function.

\begin{aligned} \mathrm P(X=3)&=\binom63(0.3)^3(0.7)^3 && \text{three of each outcome} \\ &=20(0.3)^3(0.7)^3 && \text{choose three positions} \\ &=0.18522 && \text{calculator evaluation} \end{aligned}

Why this step? There are $20$ arrangements of three passes, not the $15$ arrangements used for two passes.

Answer. $\mathrm P(X=3)=0.18522$ .

Check. Dividing this probability by the previous one gives $4/7$ ; the ratio also follows by dividing the two formula expressions.

Use your calculator for numerical binomial probabilities; the formula explains what its exact function computes.

Watch out

Wrong: $\mathrm P(X=3)=(0.3)^3(0.7)^3$ . This counts one arrangement. Right: multiply by $\binom63$ to count all possible positions of the three passes.

In the next check, define the counted category before entering the parameters.

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N1, N3 - Discrete and binomial distributions

Discrete probability distributions

Binomial conditions and exact probabilities

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