What does the column graph of a binomial distribution show?

Each column sits at a value x and its height is P(X=x), so the columns display the whole distribution and always add to 1. The tallest column marks the most likely number of successes. When p equals 0.5 the graph is symmetric; when p is less than 0.5 it leans right, and when p is greater than 0.5 it leans left.

What does the column graph of a binomial distribution show?

Each column sits at a value x and its height is P(X=x), so the columns display the whole distribution and always add to 1. The tallest column marks the most likely number of successes. When p equals 0.5 the graph is symmetric; when p is less than 0.5 it leans right, and when p is greater than 0.5 it leans left.

Year 12 Methods (Unit 3 & 4) Bernoulli sequences and the binomial distribution

Introduction to Bernoulli sequences and the binomial distribution

Q: What is a Bernoulli trial?

A Bernoulli trial is a single trial with exactly two possible outcomes, usually called success and failure, with a fixed probability p of success. Tossing a coin and recording heads or tails, or testing one item and recording pass or fail, are Bernoulli trials.

Q: When does a random variable have a binomial distribution?

X has a binomial distribution B(n,p) when there is a fixed number n of independent Bernoulli trials, each with the same success probability p, and X counts the number of successes. All four conditions must hold: two outcomes, fixed n, independent trials, and constant p.

Q: What is the binomial probability formula?

For X ~ B(n,p) the probability of exactly x successes is P(X=x) equals n choose x, times p to the power x, times (1-p) to the power (n-x), for x = 0, 1, up to n. The n choose x factor counts the arrangements of the x successes.

Q: How do you find at least and at most probabilities?

At most k means P(X is less than or equal to k); at least k means P(X is greater than or equal to k) which equals 1 minus P(X is less than or equal to k-1); between a and b inclusive means P(X is greater than or equal to a and less than or equal to b). These cumulative probabilities are found with QCAA-approved technology, such as binomcdf on a calculator.

Q: Why is sampling without replacement not binomial?

Because removing each item changes the make-up of what is left, so the success probability p is not the same on every draw and the trials are not independent. This breaks two of the binomial conditions, so the count is not binomial unless the population is very large.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a Bernoulli trial is a single trial with two outcomes — success or failure — with success probability \(p\). A Bernoulli sequence is \(n\) independent such trials, all with the same \(p\); the number of successes \(X\) is a binomial random variable \(X\sim B(n,p)\), with probability function \(P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\).

A Bernoulli trial is a single trial with exactly two outcomes, a success (probability \(p\)) or a failure (probability \(1-p\)) — a coin toss, or testing whether one item is defective.

A Bernoulli sequence is a run of \(n\) Bernoulli trials that are independent and all have the same success probability \(p\). The number of successes in the sequence, \(X\), is a binomial random variable, written

\(X\sim B(n,p),\)

with parameters \(n\) (the number of trials) and \(p\) (the success probability). \(X\) can take the values \(0,1,2,\dots,n\), and its probability function is \(P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\). The binomial model applies only when there are two outcomes, a fixed \(n\), independent trials, and a constant \(p\).

Key idea. Two outcomes + fixed \(n\) + independent trials + constant \(p\) \(\Rightarrow\) the count of successes \(X\sim B(n,p)\), with \(P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\).

\(B(6,0.4)\): each column is \(P(X=x)\); tallest at \(x=2\), leaning right (\(p<\tfrac12\))

\(B(5,0.5)\): symmetric, with equal tallest columns at \(x=2\) and \(x=3\) (\(p=\tfrac12\))

The binomial probability function, for \(X\sim B(n,p)\) and \(x=0,1,\dots,n\):

\[P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\]

P (X = x) = (\binom{n}{x}) p^{x} {(1 - p)}^{n - x}

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

\[\dbinom{n}{x}=\dfrac{n!}{x!\,(n-x)!}\]

(\binom{n}{x}) = \frac{n!}{x! (n - x)!}

Cumulative and interval probabilities (found with technology):

\[P(X\le k),\quad P(X\ge k)=1-P(X\le k-1),\quad P(a\le X\le b)=P(X\le b)-P(X\le a-1)\]

Language of probability. “At most \(k\)” is \(P(X\le k)\); “at least \(k\)” is \(P(X\ge k)\); “more than \(k\)” is \(P(X\ge k+1)\); “fewer than \(k\)” is \(P(X\le k-1)\); “between \(a\) and \(b\) inclusive” is \(P(a\le X\le b)\).

How to work with a binomial distribution

Check the conditions. Two outcomes per trial, a fixed number \(n\) of trials, independent trials, and a constant success probability \(p\). If all hold, \(X\sim B(n,p)\).
Identify \(n\), \(p\) and \(x\). \(n\) is how many trials, \(p\) is the success probability, and \(x\) is the number of successes you want.
Apply the formula. For a single value, \(P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\).
Translate the language. Rewrite “at least”, “at most”, “more than” or “between” as a range of \(x\)-values, then add the terms or use cumulative technology (e.g. \(\texttt{binomcdf}\)).

Technology. QCAA-approved calculators give \(P(X=x)\) with \(\texttt{binompdf}(n,p,x)\) and \(P(X\le k)\) with \(\texttt{binomcdf}(n,p,k)\). State the set-up, then write the value to the required number of decimal places.

Example 1 — Is it binomial?

A fair coin is tossed \(20\) times. Let \(X\) be the number of heads. Is \(X\) binomial?

Solution

Check the four conditions: two outcomes (head/tail), \(n=20\) fixed, tosses independent, and \(p=\tfrac12\) constant.

All four hold, so \(X\sim B(20,0.5)\) — yes, binomial.

X ~ B (20, 0.5)

Example 2 — A single \(P(X=x)\)

\(X\sim B(6,0.4)\). Find \(P(X=2)\), to \(4\) decimal places.

Solution

Use the formula with \(n=6\), \(x=2\).

\(P(X=2)\)	\(=\)	\(\dbinom{6}{2}(0.4)^{2}(0.6)^{4}\)
	\(=\)	\(15\times 0.16\times 0.1296\)
	\(=\)	\(0.3110\)

P (X = 2) \approx 0.3110

Example 3 — At least one

A die is rolled \(5\) times. Find \(P(\text{at least one six})\).

Solution

Here \(X\sim B(5,\tfrac16)\). Use the complement of “no sixes”.

\(P(X\ge 1)\)	\(=\)	\(1-P(X=0)\)
	\(=\)	\(1-\left(\dfrac{5}{6}\right)^{5}\)
	\(=\)	\(\dfrac{4651}{7776}\approx 0.598\)

P (X \geq 1) \approx 0.598

Example 4 — Between (inclusive)

\(X\sim B(10,0.4)\). Find \(P(3\le X\le 5)\), to \(4\) decimal places.

Solution

An inclusive range: keep \(x=3,4,5\). Use a difference of cumulative values from technology.

\(P(3\le X\le 5)\)	\(=\)	\(P(X\le 5)-P(X\le 2)\)
	\(=\)	\(0.8338-0.1673\)
	\(=\)	\(0.6665\)

P (3 \leq X \leq 5) \approx 0.6665

Common pitfalls

Do not drop the coefficient \(\dbinom{n}{x}\). The term \(p^{x}(1-p)^{n-x}\) is the probability of one particular order of successes and failures; \(\dbinom{n}{x}\) counts all the orders, so it must stay.

Without replacement is not binomial. If items are drawn and not replaced, \(p\) changes each draw and the trials are not independent — the binomial model does not apply (unless the population is very large).

Match the inequality to the words. “More than \(3\)” is \(P(X\ge 4)\), not \(P(X\ge 3)\); “at least \(3\)” is \(P(X\ge 3)\). An off-by-one here is the most common error in these questions.

Frequently asked questions

What is a Bernoulli trial?

A single trial with exactly two outcomes — a success (probability \(p\)) or a failure (probability \(1-p\)). A coin toss or a single pass/fail test is a Bernoulli trial.

When does X have a binomial distribution?

When there is a fixed number \(n\) of independent trials, each a Bernoulli trial with the same success probability \(p\), and \(X\) counts the successes. Then \(X\sim B(n,p)\).

What is the binomial probability formula?

\(P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}\) for \(x=0,1,\dots,n\). The \(\dbinom{n}{x}\) factor counts the arrangements of the \(x\) successes.

How do you find “at least” and “at most” probabilities?

“At most \(k\)” is \(P(X\le k)\); “at least \(k\)” is \(P(X\ge k)=1-P(X\le k-1)\). Use cumulative technology such as \(\texttt{binomcdf}\).

Why is sampling without replacement not binomial?

Removing each item changes what is left, so \(p\) is not constant and the trials are not independent. That breaks the binomial conditions.

What does the column graph show?

Each column's height is \(P(X=x)\), and the columns add to \(1\). The tallest column is the most likely number of successes; \(p=\tfrac12\) gives a symmetric graph.

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