9.5 Waiting-time problems

In Module 9.4, we fixed the number of trials and counted successes. Now reverse the question: keep trying until the first success and ask how long it takes. This is a waiting-time problem. The answer depends on whether each attempt still has the same success chance after previous failures.

This is intermediate probability. OpenStax's geometric-distribution lesson covers independent first-success trials; revisit its binomial lesson to compare the two questions. Conditional probability is needed when an attempt changes what remains.

A basic idea: failures before one success

Suppose each independent repair attempt succeeds with chance $1/4$. To succeed for the first time on attempt 3, the first two must fail and the third must succeed: $(3/4)(3/4)(1/4)=9/64$. It is important that “first on attempt 3” excludes a success on attempts 1 or 2.

A quick tour of waiting-time rules

Let $T$ be the number of trials including the first success. If trials are independent and every trial succeeds with the same probability $p>0$, then $T$ has a geometric distribution:

$$P(T=k)=(1-p)^{k-1}p\quad(k=1,2,\ldots), \qquad P(T>k)=(1-p)^k.$$

The first expression needs $k-1$ failures then one success. The second needs failures on all of the first $k$ trials. The expected wait is $E[T]=1/p$. One way to see it is to let $E$ be the expected number of trials: one trial is used immediately; with probability $1-p$ it fails and the process restarts with the same expected wait. Thus $E=1+(1-p)E$, giving $E=1/p$.

If trials are without replacement, the chance of success usually changes after each failure. Then multiply the actual conditional branch probabilities or enumerate possible stopping times; do not automatically use the geometric formula. Also check whether the question counts trials until success or failures before success—the latter is $T-1$.

Application 1: repeated repair attempts

Independent repair attempts each succeed with probability $1/4$. Find the probability that the first success is on attempt $3$, and the expected number of attempts.

1. Attempts 1 and 2 fail with probability $3/4$ each, then attempt 3 succeeds with probability $1/4$.
2. $P(T=3)=(3/4)^2(1/4)=9/64$.
3. $E[T]=1/(1/4)=4$ attempts. This average is not a promise that success occurs by attempt 4.

Application 2: drawing without replacement

A bag contains three red and two blue balls. Draw without replacement until the first red. What is the expected number of draws?

1. $P(T=1)=3/5$. To stop at $2$, draw blue then red: $P(T=2)=(2/5)(3/4)=3/10$.
2. To stop at $3$, draw both blues first: $P(T=3)=(2/5)(1/4)=1/10$. These probabilities total $1$.
3. $E[T]=1(3/5)+2(3/10)+3(1/10)=3/2$ draws. Using $1/(3/5)=5/3$ would be wrong because the success probability changes after a blue draw.