Independent event sequences
Independent trials have no change in the probability of one trial caused by another. Multiply the probability of every required success or failure, and use a complement for an at-least-one event
\[\Pr(E_1\cap E_2)=\Pr(E_1)\Pr(E_2),\quad\Pr(\text{at least one})=1-\Pr(\text{none})\]
- Multiply a specified pattern A pattern that requires an outcome at each trial uses one factor for each required result \(\displaystyle \Pr(S,F,S)=p_1(1-p_2)p_3\)
- Use the failure complement The probability of failure on a trial is one minus its success probability \(\displaystyle \Pr(E_i^{\prime})=1-p_i\)
- Use a complement for at least one At least one success is the complement of failure on every trial \(\displaystyle \Pr(\text{at least one})=1-(1-p)^n\)
- Use each trial's probability When success probabilities differ, use the probability belonging to each trial \(\displaystyle \Pr(S,F)=\frac{2}{3}\times\frac{1}{4}=\frac{1}{6}\)
Worked example
Four independent trials have success probabilities one half, two thirds, three quarters and one fifth. Find the probability of at least one success
\[\Pr(\text{at least one success})\]
- \(\displaystyle \Pr(E_1^{\prime})=\frac{1}{2},\quad\Pr(E_2^{\prime})=\frac{1}{3},\quad\Pr(E_3^{\prime})=\frac{1}{4},\quad\Pr(E_4^{\prime})=\frac{4}{5}\)Subtract each success probability from 1 to find the corresponding failure probability
- \(\displaystyle \Pr(\text{no successes})=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}\times\frac{4}{5}=\frac{1}{30}\)Independence means that the four failure probabilities are multiplied
- \(\displaystyle \Pr(\text{at least one success})=1-\Pr(\text{no successes})=1-\frac{1}{30}=\frac{29}{30}\)Subtract the probability of no successes from 1.
Common mistakes
Adding overlapping success events
✕\(\displaystyle \Pr(A)=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)✓\(\displaystyle \Pr(A)=1-\left(\frac{1}{2}\right)^3=\frac{7}{8}\)
Suppose three independent trials each have success probability one half, and A means at least one success. The individual success events overlap, so adding them directly counts some outcomes more than onceOmitting required failure factors
✕\(\displaystyle \Pr(S,F,S)=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)✓\(\displaystyle \Pr(S,F,S)=\frac{2}{3}\times\frac{1}{2}\times\frac{2}{3}=\frac{2}{9}\)
Suppose the three independent trials have success probabilities two thirds, one half and two thirds. A required failure contributes its complement probability and must be included in the productUsing the complement as the answer
✕\(\displaystyle \Pr(A)=\Pr(A^{\prime})=\frac{1}{8}\)✓\(\displaystyle \Pr(A)=1-\Pr(A^{\prime})=1-\frac{1}{8}=\frac{7}{8}\)
Suppose three independent trials each have success probability one half, and A means at least one success. The probability of at least one success is one minus the probability of no successesDifficulty levels
introCalculate an all-success, no-success or specified success-failure pattern with a common probability
standardCalculate an only-subset or at-least-one event, often using a complement
challengeUse distinct probabilities for several trials and combine disjoint first-success patterns
examChoose a suitable strategy and calculate one exact independent-event probability
Questions about this topic
Why are probabilities multiplied for independent trials?+
Independence means that the required result on one trial does not change the probability on another trial. The joint pattern probability is therefore the product of its factors
How is the probability of at least one success found?+
Find the probability of failure on every trial, then subtract it from 1. Failure on every trial is the complement of at least one success
Why must failure factors be included in an only-success pattern?+
Only the named trials may succeed, so every other trial must fail. Each required failure contributes a factor of 1 minus its success probability
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