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Year 11 Maths - Methods (Unit 1 and Unit 2) Discrete probability distributions

Sampling With Replacement - The Binomial Distribution

20 practice questions 0 video lessons Theory + worked examples

Meet the binomial distribution for Victorian Year 11 Mathematical Methods (VCAA) — the pattern of probabilities for the number of successes in a fixed number of independent trials.

You will learn to recognise when a situation is binomial, identify the number of trials and the success probability, and calculate probabilities for exactly, at most and at least a given number of successes.

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Theory

In Year 11 Mathematical Methods (Unit 2), a binomial distribution describes the number of successes \(X\) in a fixed number of independent trials, each with the same success probability. We write \(X\sim\text{Bi}(n,\,p)\). This page sets out the binomial conditions, the probability formula \(P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}\), and how to find exact, “at least” and cumulative probabilities.

A binomial experiment is a fixed number \(n\) of trials that each meet four conditions:

  • there are exactly two outcomes each trial — a success or a failure;
  • the trials are independent (one outcome does not affect the next);
  • the success probability \(p\) is constant across all trials;
  • the number of trials \(n\) is fixed in advance.

Let \(X\) be the number of successes in the \(n\) trials. Then \(X\) has a binomial distribution, written \(X\sim\text{Bi}(n,\,p)\), and \(X\) can take the whole-number values \(0,1,2,\dots,n\). The failure probability is \(1-p\), sometimes written \(q\).

Same setting every time. Repeated, replaceable trials — tossing a coin, spinning a spinner, sampling from a large batch — keep \(p\) fixed and the trials independent, so \(X\) counts successes in a binomial experiment.
Binomial histogram Bi(6, 0.25)Probability histogram for X binomial with six trials and success probability 0.25, skewed to the right. 0 1 2 3 4 5 6 k P(X=k)
\(X\sim\text{Bi}(6,\,0.25)\): with \(p<0.5\) the histogram is skewed right; the tallest bar sits at a low \(k\).
Binomial histogram Bi(8, 0.5)Symmetric probability histogram for X binomial with eight trials and success probability one half. 0 1 2 3 4 5 6 7 8 k P(X=k)
\(X\sim\text{Bi}(8,\,0.5)\): with \(p=0.5\) the histogram is symmetric about \(k=\tfrac{n}{2}=4\).

For \(X\sim\text{Bi}(n,\,p)\), the probability of exactly \(k\) successes is:

\[P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k},\qquad k=0,1,\dots,n\]
P(X=k)=(nk)pk(1-p)n-k

where the binomial coefficient counts the arrangements of the \(k\) successes:

\[\binom{n}{k}=\dfrac{n!}{k!\,(n-k)!}\]

For “at least one” use the complement, and for a range add the terms:

\[P(X\ge 1)=1-P(X=0),\qquad P(X\le 1)=P(X=0)+P(X=1)\]
P(X≥1)=1-P(X=0)
Every term has three parts: the count \(\binom{n}{k}\), the successes \(p^{k}\), and the failures \((1-p)^{n-k}\). The powers \(k\) and \(n-k\) must always add back to \(n\).

How to find a binomial probability

  1. Check it is binomial: fixed \(n\) trials, two outcomes, independent, constant \(p\).
  2. Identify \(n\), \(p\), \(k\): \(n\) is the number of trials, \(p\) the success probability on one trial, \(k\) the number of successes asked for.
  3. Apply the formula: substitute into \(P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}\) and evaluate (usually to \(4\) decimal places).
  4. Handle words: “at least one” \(\Rightarrow 1-P(X=0)\); “at most” or “fewer than” \(\Rightarrow\) add the separate terms.
Example 1 — Exactly k successes
A die is rolled \(6\) times. A “success” is rolling more than \(4\), with probability \(0.25\) each roll (rolls independent). Let \(X\sim\text{Bi}(6,\,0.25)\). Find \(P(X=2)\).
Solution

Identify the parameters:

\(n\)\(=\)\(6\)
\(p\)\(=\)\(0.25\)
\(k\)\(=\)\(2\)

Substitute into \(P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}\):

\(P(X=2)\)\(=\)\(\binom{6}{2}(0.25)^{2}(0.75)^{4}\)

Evaluate each factor, then multiply:

\(=\)\(15\times 0.0625\times 0.316406\ldots\)
\(=\)\(0.2966\)

\(P(X=2)=0.2966\).

Bi(6, 0.25) with k equals 2 highlightedHistogram of X binomial six trials probability 0.25, with the bar at two greater than four highlighted. 0 1 2 3 4 5 6 k P(X=k)
P(X=2)=0.2966
Example 2 — “At least one” (complement)
A seed germinates with probability \(0.3\). Five seeds are planted independently, \(X\sim\text{Bi}(5,\,0.3)\). Find \(P(X\ge 1)\), the probability at least one germinates.
Solution

“At least one” is easiest by the complement:

\(P(X\ge 1)\)\(=\)\(1-P(X=0)\)

Find \(P(X=0)\) — no successes, so all five fail:

\(P(X=0)\)\(=\)\(\binom{5}{0}(0.3)^{0}(0.7)^{5}\)
\(=\)\(0.16807\)

Subtract from \(1\):

\(P(X\ge 1)\)\(=\)\(1-0.16807\)
\(=\)\(0.8319\)

\(P(X\ge 1)=0.8319\).

Bi(5, 0.3) with at least one highlightedHistogram of X binomial five trials probability 0.3, with every bar from one upward highlighted for at least one success. 0 1 2 3 4 5 k P(X=k)
P(X≥1)=0.8319
Example 3 — Cumulative “at most”
In a large batch, \(20\%\) of globes are energy-saving, so \(p=0.2\) stays constant. Six are chosen, \(X\sim\text{Bi}(6,\,0.2)\). Find \(P(X\le 1)\).
Solution

“At most one” means \(0\) or \(1\) success:

\(P(X\le 1)\)\(=\)\(P(X=0)+P(X=1)\)

Compute each term separately:

\(P(X=0)\)\(=\)\((0.8)^{6}=0.262144\)
\(P(X=1)\)\(=\)\(\binom{6}{1}(0.2)(0.8)^{5}=0.393216\)

Add the two probabilities:

\(P(X\le 1)\)\(=\)\(0.262144+0.393216\)
\(=\)\(0.6554\)

\(P(X\le 1)=0.6554\).

Bi(6, 0.2) with k equals 0 and 1 highlightedHistogram of X binomial six trials probability 0.2, with the bars at zero and one highlighted for at most one. 0 1 2 3 4 5 6 k P(X=k)
P(X≤1)=0.6554
Example 4 — Two-part: exact and a tail
A fair coin is tossed \(8\) times, \(X\sim\text{Bi}(8,\,0.5)\). Find (i) \(P(X=4)\) and (ii) \(P(X\ge 6)\), the probability of at least six heads.
Solution

(i) Exactly four heads — substitute \(k=4\):

\(P(X=4)\)\(=\)\(\binom{8}{4}(0.5)^{4}(0.5)^{4}\)
\(=\)\(70\times (0.5)^{8}\)
\(=\)\(0.2734\)

(ii) “At least six” sums the upper tail \(k=6,7,8\):

\(P(X\ge 6)\)\(=\)\(\big(\tbinom{8}{6}+\tbinom{8}{7}+\tbinom{8}{8}\big)(0.5)^{8}\)
\(=\)\((28+8+1)\times 0.00390625\)
\(=\)\(0.1445\)

\(P(X=4)=0.2734\) and \(P(X\ge 6)=0.1445\).

Bi(8, 0.5) with the upper tail highlightedSymmetric histogram of X binomial eight trials probability one half, with the bars at six, seven and eight highlighted. 0 1 2 3 4 5 6 7 8 k P(X=k)
P(X=4)=0.2734

Common pitfalls

Mixing up \(n\), \(k\) and \(p\). \(n\) is how many trials, \(k\) is how many successes you want, and \(p\) is the chance of success on one trial — keep them straight before substituting.
Forgetting the failure factor. Every term needs \((1-p)^{n-k}\) as well as \(p^{k}\); the two powers \(k\) and \(n-k\) must add back to \(n\).
Using \(p\) for the wrong outcome. \(p\) is the probability of the event you are counting. If you are counting late buses, \(p\) is the chance of being late, not on time.
Adding instead of using the complement. For “at least one” do \(1-P(X=0)\) rather than summing many terms — it is faster and less error-prone.
Not checking the binomial conditions. If \(p\) changes between trials, or trials are not independent (e.g. sampling a small group without replacement), the distribution is not binomial.

Frequently asked questions

When is a distribution binomial?

When there is a fixed number \(n\) of independent trials, each with only two outcomes and the same success probability \(p\), and \(X\) counts the successes. Then \(X\sim\text{Bi}(n,\,p)\).

What is the binomial probability formula?

\(P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}\), the number of arrangements \(\binom{n}{k}\) times \(k\) successes \(p^{k}\) times \(n-k\) failures \((1-p)^{n-k}\).

How do I find “at least one”?

Use the complement: \(P(X\ge 1)=1-P(X=0)\), where \(P(X=0)=(1-p)^{n}\). This avoids adding every other term.

What do n, p and k mean?

\(n\) is the number of trials, \(p\) is the probability of success on a single trial, and \(k\) is the number of successes whose probability you want.

What is the shape of a binomial distribution?

The histogram is symmetric when \(p=0.5\), skewed right when \(p<0.5\), and skewed left when \(p>0.5\).

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