Search topics

Search

Exact probability of repeated independent successes

Start a set of 8
Step 2 of 5

Exact binomial probability

A binomial variable counts successes in a fixed number of independent trials with the same success probability. For an exact success count, count the possible positions of the successes and include the complementary failures

\[\Pr(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\]
  1. 1Use the combination coefficient The coefficient counts which trial positions contain the successes \(\displaystyle \binom{6}{2}=15\)
  2. 2Use the complementary probability The failure probability is one minus the success probability \(\displaystyle p=\frac{2}{5},\quad 1-p=\frac{3}{5}\)
  3. 3Make the exponents total n The success and failure exponents together count every trial \(\displaystyle 2 + (6-2) = 6\)
  4. 4Add disjoint success counts For a cumulative event, add the exact terms for the allowed success counts, or use a complement when it is shorter \(\displaystyle \Pr(X\ge 2)=\Pr(X=2)+\Pr(X=3)+\cdots\)
Year 11, VCE Units 1 & 2, Chapter 8, 1 difficulty level, Calculator allowedBuild a worksheet

Exact binomial probability

A binomial variable counts successes in a fixed number of independent trials with the same success probability. For an exact success count, count the possible positions of the successes and include the complementary failures

\[\Pr(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\]
  1. Use the combination coefficient The coefficient counts which trial positions contain the successes \(\displaystyle \binom{6}{2}=15\)
  2. Use the complementary probability The failure probability is one minus the success probability \(\displaystyle p=\frac{2}{5},\quad 1-p=\frac{3}{5}\)
  3. Make the exponents total n The success and failure exponents together count every trial \(\displaystyle 2 + (6-2) = 6\)
  4. Add disjoint success counts For a cumulative event, add the exact terms for the allowed success counts, or use a complement when it is shorter \(\displaystyle \Pr(X\ge 2)=\Pr(X=2)+\Pr(X=3)+\cdots\)

Worked example

A repeated independent trial has a success probability of one-third and is performed 6 times. Find the probability of exactly 2 successes

\[\Pr(X=2)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4\]
  1. \(\displaystyle p=\frac{1}{3},\quad 1-p=\frac{2}{3},\quad n=6,\quad r=2\)Identify the success probability, failure probability, trial count and success count
  2. \(\displaystyle \binom{6}{2}=\frac{6\times 5}{2\times 1}=15\)Count the positions that can contain the 2 successes
  3. \(\displaystyle \Pr(X=2)=15\times\left(\frac{1}{3}\right)^2\times\left(\frac{2}{3}\right)^4\)Use the binomial term with 2 successes and 4 failures
  4. \(\displaystyle 15\times\frac{1}{9}\times\frac{16}{81}=\frac{240}{729}=\frac{80}{243}\)Multiply the exact factors and reduce the fraction

Common mistakes

Omitting the combination coefficient
✕\(\displaystyle \Pr(X=2)=\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3=\frac{1}{32}\)✓\(\displaystyle \Pr(X=2)=\binom{5}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3=\frac{5}{16}\)
The combination coefficient counts the possible positions of the successes, so it cannot be omitted
Using equal success and failure exponents
✕\(\displaystyle \binom{5}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2=\frac{5}{8}\)✓\(\displaystyle \binom{5}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3=\frac{5}{16}\)
The failure exponent is the remaining number of trials, so the two exponents must add to 5
Multiplying cumulative terms
✕\(\displaystyle \Pr(X\ge 2)=\Pr(X=2)\times\Pr(X=3)\times\Pr(X=4)\times\Pr(X=5)\)✓\(\displaystyle \Pr(X\ge 2)=\Pr(X=2)+\Pr(X=3)+\Pr(X=4)+\Pr(X=5)\)
Different success counts are disjoint alternatives, so their probabilities are added

Difficulty levels

challengeCumulative or range probabilities with a direct or complement strategy

Questions about this topic

Why is the combination coefficient needed?+

A success count does not specify which trials were successful. The combination coefficient counts all possible positions of those successes

Why is the failure exponent the difference between the trial count and success count?+

Subtract the success count from the total trial count to get the number of failures

When is a complement useful for a cumulative probability?+

Use a complement when the success counts outside the requested event form a shorter set of terms. Subtract their exact total from 1

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 8: Probability and counting

In this chapter

Finite events, probability assignments and complementsRepresent a multi-stage finite sample spaceCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConditional probability from exact representationsDistribution of a finite random variableDesign and interpret a probability simulationCount restricted selections and calculate exact probabilitiesCount arrangements with product, sum and restriction rulesTest exact independenceExact probability of repeated independent successesMulti-stage probability, multiplication and total probabilityProbability from a uniform area modelProbabilities of sequences of independent events