Probability distribution Revision Notes for AQA GCSE Statistics

Probability distribution

What is a probability distribution?

A probability distribution shows us all the possible values that a random variable can take, along with the probability of each value occurring. Think of it as a complete list of what could happen and how likely each outcome is.

The outcomes in a probability distribution must be numerical. For example, when we toss three fair coins, we can count the number of heads we get - this could be 0, 1, 2, or 3 heads.

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Worked Example: Three Fair Coins

Let's look at this coin example:

X represents the number of heads when tossing three fair coins
X can equal 0, 1, 2, or 3
Each outcome has a specific probability:
- P(X = 0) = 0.125 (getting no heads)
- P(X = 1) = 0.375 (getting exactly one head)
- P(X = 2) = 0.375 (getting exactly two heads)
- P(X = 3) = 0.125 (getting three heads)

Notice that all these probabilities add up to 1, which they always must in any probability distribution.

Binomial distribution

A binomial distribution is a special type of probability distribution that occurs when an experiment has exactly two possible outcomes. We call one outcome a success and the other a failure.

Key characteristics of binomial distributions

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Essential Conditions for Binomial Distribution

For a situation to follow a binomial distribution, it must meet these conditions:

There are only two possible outcomes for each trial
The probability of success remains the same for each trial
Each trial is independent of the others
We have a fixed number of trials

Understanding success and failure

Let's say we're rolling a dice and counting getting a 6 as a success. Then:

Getting a 6 = success (let's call this probability p)
Getting 1, 2, 3, 4, or 5 = failure (let's call this probability q)

Since these are the only two possibilities: $p + q = 1$

This means if we know p, we can always find q by calculating $q = 1 - p$ .

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Understanding the relationship between p and q is crucial - they are complementary probabilities that must always sum to 1 in any binomial situation.

Probability trees for binomial situations

When we have two trials of a binomial situation, we can use a probability tree diagram to show all possible outcomes. For our dice example with two throws:

First throw: probability p of getting 6, probability q of not getting 6
Second throw: same probabilities, but independent of the first

The tree shows us four possible paths:

Success then success: probability $p \times p = p^2$
Success then failure: probability $p \times q = pq$
Failure then success: probability $q \times p = qp$
Failure then failure: probability $q \times q = q^2$

Calculating probabilities with the binomial expansion

For two trials of a binomial distribution, we can use the expansion:

$(p + q)^2 = p^2 + 2pq + q^2$

This formula tells us:

$p^2$ = probability of two successes
$2pq$ = probability of exactly one success (this can happen in two different ways)
$q^2$ = probability of two failures

This expansion works because we're considering all possible combinations of successes and failures across our two trials.

Worked example: bags and beads

Let's work through a practical example to see how these concepts apply.

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Worked Example: Bags and Beads

Problem: Two bags A and B each contain 5 white beads and 3 red beads. One bead is taken randomly from each bag.

Part (a): Find the probability that both beads are white

First, let's identify our success: getting a white bead.

Each bag has 8 beads total (5 white + 3 red)
Probability of white from each bag: $p = \frac{5}{8}$
Probability of red from each bag: $q = \frac{3}{8}$

For both beads to be white, we need success from both bags: Probability = $p^2 = \left(\frac{5}{8}\right)^2 = \frac{25}{64}$

Part (b): Find the probability that exactly one bead is white

This means one white bead and one red bead, which can happen in two ways:

White from bag A, red from bag B: $\frac{5}{8} \times \frac{3}{8} = \frac{15}{64}$
Red from bag A, white from bag B: $\frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$

Total probability = $\frac{15}{64} + \frac{15}{64} = \frac{30}{64}$

Alternatively, using our formula: $2pq = 2 \times \frac{5}{8} \times \frac{3}{8} = \frac{30}{64}$

Practice problem: biassed dice

Here's a problem to test your understanding:

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Practice Problem: Biassed Dice

Alice has a biassed dice where the probability of rolling a 6 is 0.2. She rolls the dice twice.

(a) Work out the probability that she gets exactly one 6

$p = 0.2$ (probability of success - getting a 6)
$q = 1 - 0.2 = 0.8$ (probability of failure - not getting a 6)
Probability of exactly one success = $2pq = 2 \times 0.2 \times 0.8 = 0.32$

(b) Work out the probability that she does not get a 6 in either roll

This means two failures: $q^2 = (0.8)^2 = 0.64$

Common exam tips

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Helpful Exam Strategies

When tackling binomial distribution problems:

Always identify what counts as success and failure
Check that $p + q = 1$
Remember that "exactly one success in two trials" equals $2pq$
Use probability trees if you're unsure - they help visualise all outcomes
Double-check that all probabilities in your distribution add up to 1

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Key Points to Remember:

A probability distribution shows all possible numerical outcomes and their probabilities
Binomial distributions have exactly two outcomes: success and failure
The probabilities of success (p) and failure (q) must add up to 1: $p + q = 1$
For two trials: probability of exactly one success = $2pq$ , two successes = $p^2$ , no successes = $q^2$
Always check your conditions: fixed number of trials, independent outcomes, same probability each time

Probability distribution (AQA GCSE Statistics): Revision Notes