The geometric distributionEdexcel A-Level Further Maths: Revision notes

Suppose you carry out independent trials, each with the same probability $p$ of success, and count how many trials it takes to get the first success. This number is a geometric random variable $X$, written $X\sim\mathrm{Geo}(p)$. The model needs three conditions: each trial has only two outcomes (success or failure); the probability of success $p$ is the same every time; and the trials are independent. $X$ counts the trial on which the success happens, so it can take the values $x=1,2,3,\dots$ with no upper limit. For example, the number of rolls of a fair die up to and including the first six is $\mathrm{Geo}\left(\frac16\right)$.

The first success is on trial $x$ when the first $x-1$ trials all fail and trial $x$ succeeds. Writing $q=1-p$, $\mathrm{P}(X=x)=p(1-p)^{x-1},\quad x=1,2,3,\dots$ The tail probabilities are simple, because $X>x$ just means the first $x$ trials all fail: $\mathrm{P}(X>x)=(1-p)^x,\qquad \mathrm{P}(X\geq x)=(1-p)^{x-1},\qquad \mathrm{P}(X\leq x)=1-(1-p)^x.$ Worked example: a component is defective with probability $0.08$. $\mathrm{P}(X=4)=0.92^3\times0.08=0.0623$ and $\mathrm{P}(X>5)=0.92^5=0.659$. For the smallest $n$ with $\mathrm{P}(X\leq n)>0.5$, solve $0.92^n<0.5$ using logarithms: $n>8.31$, so $n=9$. Remember to reverse the inequality when dividing by the negative number $\ln0.92$.

For $X\sim\mathrm{Geo}(p)$: $\mu=\mathrm{E}(X)=\frac1p,\qquad \sigma^2=\mathrm{Var}(X)=\frac{1-p}{p^2}.$ Both are in the formulae booklet, and you do not have to prove them. A smaller $p$ makes success rarer, so the mean wait and the spread both grow. Worked example: for the die, $p=\frac16$, giving $\mu=6$ rolls and $\sigma^2=\frac{5/6}{1/36}=30$, so $\sigma=5.48$. If you are told $\mathrm{E}(X)=5$, then $p=\frac15$ and $\mathrm{Var}(X)=\frac{0.8}{0.04}=20$. The standard deviation is the square root of the variance.

In context questions, say what counts as a trial and a success, then state $X\sim\mathrm{Geo}(p)$ with its value of $p$. Typical tasks are to find a probability for a given $x$, to find the smallest $n$ for a target probability, and to use a series for probabilities such as $\mathrm{P}(X\text{ is odd})$, where the terms $p,\ p(1-p)^2,\ p(1-p)^4,\dots$ form a geometric series with ratio $(1-p)^2$. The sum to infinity $\frac{a}{1-r}$ then gives $\mathrm{P}(X\text{ odd})=\frac{p}{1-(1-p)^2}=\frac{1}{2-p}$. In a game where two players alternate and the first to succeed wins, the player who goes first wins when $X$ is odd. Assumptions to comment on: the trials must be independent and $p$ must stay constant. If the probability changes, for example because a player improves with practice, the geometric model is not suitable.