Binomial event probabilities
A binomial event is found by identifying the integer success counts that satisfy its condition. Add the exact probabilities for those counts, or use a complement when it gives a shorter calculation
\[\Pr(X=x)=\binom{n}{x}(1-p)^{n-x}p^x\]
- Translate the event List the integer success counts included by the inequality, paying attention to strict endpoints \(\displaystyle X\le 2\Longleftrightarrow X\in\{0,1,2\}\)
- Calculate each term Use the binomial probability rule for every included success count \(\displaystyle \Pr(X=2)=\binom{4}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2\)
- Add alternatives Success counts in one event are disjoint alternatives, so add their probabilities \(\displaystyle \Pr(X\le 2)=\Pr(X=0)+\Pr(X=1)+\Pr(X=2)\)
- Use the complement or condition For at least one, subtract zero successes from one; for a conditional probability, divide the intersection by the conditioning event. The conditional formula requires the conditioning event to have positive probability \(\displaystyle \begin{gathered}\Pr(X\ge 1)=1-\Pr(X=0)\\\Pr(B)>0\\\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\end{gathered}\)
Worked example
For the binomial model shown, find the probability of at most two successes
\[X\sim\operatorname{Bi}\left(6,\frac{1}{3}\right)\]
- \(\displaystyle 1-p=1-\frac{1}{3}=\frac{2}{3}\)Find the failure probability from the complement of p.
- \(\displaystyle X\le 2\Longleftrightarrow x\in\{0,1,2\}\)The event includes the three integer success counts from zero to two
- \(\displaystyle \begin{gathered}\Pr(X=0)=\left(\frac{2}{3}\right)^6\\\Pr(X=0)=\frac{64}{729}\end{gathered}\)Zero successes means six independent failures, so multiply the failure probability six times
- \(\displaystyle \begin{gathered}\binom{6}{1}=\frac{6!}{1!\,5!}=6\\\Pr(X=1)=\binom{6}{1}\frac{1}{3}\left(\frac{2}{3}\right)^5\\\Pr(X=1)=6\times\frac{1}{3}\times\frac{32}{243}\\\Pr(X=1)=\frac{192}{729}\end{gathered}\)Choose the position of the single success, then multiply by the probability of one success and five failures
- \(\displaystyle \begin{gathered}\binom{6}{2}=\frac{6!}{2!\,4!}\\\binom{6}{2}=\frac{6\times5}{2\times1}=15\\\Pr(X=2)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4\\\Pr(X=2)=15\times\frac{1}{9}\times\frac{16}{81}\\\Pr(X=2)=\frac{240}{729}\end{gathered}\)Choose the two success positions without counting their order, then multiply by the probability of two successes and four failures
- \(\displaystyle \Pr(X\le 2)=\frac{64}{729}+\frac{192}{729}+\frac{240}{729}\)Calculate the three binomial terms and add them because the counts are disjoint alternatives
- \(\displaystyle \Pr(X\le 2)=\frac{496}{729}\)The exact cumulative probability is the sum of the three included terms
Common mistakes
Adding trial probabilities for an at-least-one event
✕\(\displaystyle \Pr(X\ge 1)=4\times\frac{1}{2}=2\)✓\(\displaystyle \Pr(X\ge 1)=1-\Pr(X=0)=1-\frac{1}{16}=\frac{15}{16}\)
An at-least-one event is best found by subtracting the probability of zero successes from 1.Multiplying cumulative terms
✕\(\displaystyle \Pr(X\le 2)=\frac{1}{16}\times\frac{1}{4}\times\frac{3}{8}=\frac{3}{512}\)✓\(\displaystyle \Pr(X\le 2)=\frac{1}{16}+\frac{1}{4}+\frac{3}{8}=\frac{11}{16}\)
The included success counts are disjoint alternatives, so their probabilities must be addedUsing the wrong conditional denominator
✕\(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(A)}\)✓\(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\)
The event after the vertical bar is the conditioning event and supplies the denominatorDifficulty levels
introCalculate a direct cumulative probability for a small binomial variable
standardFind an at-least-one probability by using the complement of zero successes
challengeCalculate a conditional probability from a non-adjacent intersection of success-count events
examChoose an event method or description and calculate the resulting exact probability
Questions about this topic
How are the counts for a cumulative event chosen?+
List the integer values of X that satisfy the inequality, including or excluding endpoints according to the inequality sign
Why are probabilities added for a cumulative event?+
Different success counts cannot occur at the same time, so they are disjoint alternatives and their probabilities are added
What is the denominator of a conditional probability?+
The denominator is the probability of the conditioning event, the event after the vertical bar
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 11: Binomial distribution
In this chapter
Bernoulli trials and the Bernoulli distribution11A Bernoulli trials and the Bernoulli distribution: Finite events, probability assignments and complementsSolve a quadratic exactly with the quadratic formula11A Bernoulli trials and the Bernoulli distribution: Round to decimal places and significant figures11B Binomial distribution and parameters: Binomial conditions, the distribution table and graph shapeMean, variance, standard deviation and parameter recovery of a binomial variableExact probability of repeated independent successesEvaluate factorial and combination expressionsSolve a linear equation11B Binomial distribution and parameters: Round to decimal places and significant figuresCumulative, at-least-one and conditional probabilities of a binomial variableSmallest trial count for an at-least-one threshold11C Binomial distribution events: Finite events, probability assignments and complementsConditional probability from exact representationsSolve and interpret a linear inequation11D Binomial shape and specified order: Binomial conditions, the distribution table and graph shapeProbabilities of sequences of independent eventsDescribe the shape of a distribution