Discrete random variables and probability distributions, the binomial distribution and its conditions, and calculating binomial probabilities.

What this dot point is asking

WJEC wants you to understand a discrete random variable and its probability distribution, to know the conditions that make the binomial distribution appropriate, and to calculate binomial probabilities for single values and cumulative ranges. The binomial is the distribution behind the hypothesis test in the next topic, so its conditions and notation must be secure.

The answer

Discrete random variables

A discrete random variable $X$ takes a countable set of values, and a probability distribution lists each value with its probability.

A discrete uniform distribution assigns equal probability to each of its values, such as the score on a fair die where each of $1$ to $6$ has probability $\tfrac{1}{6}$. A common short-answer task is to find an unknown probability in a distribution table by using the fact that the probabilities sum to $1$: if a table lists probabilities $0.1$, $0.3$, $k$ and $0.2$, then $k = 1 - 0.6 = 0.4$.

The binomial distribution and its conditions

These four conditions are a frequent short-answer question, so learn them precisely: fixed $n$, two outcomes, independence, constant $p$.

Calculating binomial probabilities

For ranges, use the cumulative function: $P(X \le r)$ is built into the calculator, and $P(X \ge r) = 1 - P(X \le r-1)$.

Examples in context

Example 1. Quality control

A machine produces components that are acceptable with probability $0.95$, independently. In a sample of $20$, the expected number acceptable is $np = 20 \times 0.95 = 19$, and the probability that all $20$ are acceptable is $(0.95)^{20} \approx 0.358$. The mean and a single binomial term together summarise the batch.

Example 2. Multiple-choice guessing

A student guesses all $10$ multiple-choice questions, each with $4$ options, so $p = 0.25$ and $X \sim B(10, 0.25)$. The probability of scoring at least $5$ is $P(X \ge 5) = 1 - P(X \le 4) \approx 0.0781$. The cumulative function turns an awkward sum into one calculator step.

Example 3. When the binomial does not apply

Drawing $5$ cards from a pack and counting the aces is not binomial, because the draws are without replacement, so the probability of an ace changes after each card. The independence and constant-probability conditions both fail, which is exactly the kind of distinction WJEC tests in a "state why the binomial model is or is not suitable" question.

Try this

Q1. State the four conditions for a binomial distribution. [2 marks]

• Cue. Fixed number of trials, two outcomes, independent trials, constant probability of success.

Q2. $X \sim B(6, 0.5)$. Find $P(X = 6)$. [2 marks]

• Cue. $\binom{6}{6}(0.5)^6 = (0.5)^6 = \dfrac{1}{64} \approx 0.0156$.

Q3. $X \sim B(15, 0.2)$. Find the mean of $X$. [1 mark]

• Cue. $E(X) = np = 15 \times 0.2 = 3$.