Geometric and negative binomial distributions

Explanation

• For independent Bernoulli trials with constant success probability $p$, a geometric variable $X$ counts the trial of the first success: $P(X=x)=p(1-p)^{x-1}$ for $x=1,2,\ldots$.
• A negative binomial variable $Y$ counts the trial of the $r$th success: $P(Y=y)=\binom{y-1}{r-1}p^r(1-p)^{y-r}$ for $y=r,r+1,\ldots$.
• The last trial is fixed as a success, while $r-1$ successes are placed among the preceding $y-1$ trials.
• Cumulative events can sometimes be recast as binomial counts.
• Examiners expect the support convention, the final-success condition and the independence/constant-$p$ modelling assumptions to be clear.

Explanation

• For a geometric variable counting the trial of the first success, $E(X)=1/p$ and $\operatorname{Var}(X)=(1-p)/p^2$. Its standard deviation is $\sqrt{1-p}/p$.
• These formulas use support $1,2,\ldots$, not the alternative convention that counts failures before success.
• Given the mean or variance, solve for a probability satisfying $0<p\leq1$ and check that the result is consistent.
• The memoryless property explains why, after any run of failures, the remaining waiting time has the same geometric model.
• Examiners expect the convention to match the question, the standard deviation to be positive, and probability events such as $X\geq k$ to count the required initial failures correctly.

Explanation

• If $X$ counts the trial on which the $r$th success occurs, then $E(X)=r/p$ and $\operatorname{Var}(X)=r(1-p)/p^2$.
• The ratio $\operatorname{Var}(X)/E(X)=(1-p)/p$ is efficient when both moments are known: it determines $p$, then the mean determines the positive integer $r$.
• These are formulas for the success-trial number, not for the number of failures before $r$ successes.
• Probabilities still use $\binom{x-1}{r-1}p^r(1-p)^{x-r}$.
• Examiners expect any recovered $r$ to be a positive integer, $p$ to lie in $(0,1]$, and cumulative boundaries to respect the minimum value $X=r$.