Tutorials › VEE Mathematical Statistics › Probability of Reaching r Successes by a Fixed Deadline

Home  /  Tutorials  /  VEE Mathematical Statistics / Geometric and Negative Binomial Models

VEE Mathematical Statistics · Geometric and Negative Binomial Models · Tutorial 123 of 880

Probability of Reaching r Successes by a Fixed Deadline

Translate a deadline for completing \(r\) successes into a binomial tail, and distinguish reaching the target by the deadline from reaching it exactly at the deadline.

Intermediate 9 min read

What You'll Learn

  • Define the completion-time event for reaching \(r\) successes by trial \(n\)
  • Convert that event to a binomial count probability
  • Calculate deadline probabilities with a direct sum or a complement
  • Distinguish completion by a deadline from completion exactly on a specified trial
  • Check when the fixed-probability independent-trial model is appropriate
  • Avoid common indexing and tail-direction errors

Why a Deadline Changes the Question

A waiting-time question can ask when a target is reached or whether it is reached soon enough. For example, a hypothetical laboratory may repeat an independent test until it records three successful results. The completion time is the trial number of the third success; a deadline question asks whether that time is no greater than a specified trial number.

The key simplification is to count successes at the deadline. Reaching the target by trial \(n\) is equivalent to having at least \(r\) successes among the first \(n\) trials. This turns a cumulative negative binomial probability into a binomial tail, which can often be easier to calculate and check. The method is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose trials are independent, each trial has success probability \(p\), and \(0<p<1\). Let \(T_r\) be the trial number on which the \(r\)th success occurs, where \(r\) is a positive integer. The event of completing by trial \(n\) is \(\{T_r\le n\}\). If \(N_n\) is the number of successes in the first \(n\) trials, then \(N_n\sim\operatorname{Binomial}(n,p)\), and \(\{T_r\le n\}=\{N_n\ge r\}\).

The Deadline Probability

For a fixed positive integer \(n\), the number of successes \(N_n\) in \(n\) independent trials with the same success probability \(p\) has the binomial probability function. Therefore, when \(1\le r\le n\),

$$ P(T_r\le n)=P(N_n\ge r) =\sum_{k=r}^{n}\binom{n}{k}p^k(1-p)^{n-k}. $$

Each term represents exactly \(k\) successes in the first \(n\) trials. Since completing the target by the deadline allows any count from \(r\) through \(n\), all those terms must be included. The complementary form is often shorter:

$$ P(T_r\le n) =1-P(N_n\le r-1) =1-\sum_{k=0}^{r-1}\binom{n}{k}p^k(1-p)^{n-k}. $$

The complement consists of outcomes with too few successes to complete the target. If \(n<r\), completion by trial \(n\) is impossible, so \(P(T_r\le n)=0\). If \(r=0\), the target is already met before any trial, so the probability is \(1\). The displayed binomial sums assume an integer deadline and a fixed success probability on independent trials. They do not apply unchanged when trial outcomes are dependent or the success probability varies from trial to trial.

Why the Binomial Connection Holds

Let \(I_j\) equal \(1\) when trial \(j\) is a success and \(0\) otherwise. The total number of successes by trial \(n\) is \(N_n=I_1+\cdots+I_n\). The \(r\)th success has occurred by then precisely when this total is at least \(r\). In event notation,

$$ \{T_r\le n\} =\{I_1+\cdots+I_n\ge r\} =\{N_n\ge r\}. $$

Independence and the common success probability imply that \(N_n\) has a binomial distribution: for any \(k\) from \(0\) to \(n\), there are \(\binom{n}{k}\) ways to choose which \(k\) trials succeed, and each such pattern has probability \(p^k(1-p)^{n-k}\). Thus

$$ P(N_n=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$

Summing this probability over \(k=r,\ldots,n\) proves the deadline formula. The argument depends on counting successes across the fixed set of first \(n\) trials; it does not require treating each possible completion time separately.

Worked Deadline Calculations

Worked Example: Three Successful Calibration Runs by Trial Five

A hypothetical device calibration run succeeds with probability \(p=0.4\), independently from run to run. Let \(T_3\) be the run on which the third success occurs. Find the probability of recording at least three successes by the end of five runs.

Method. Use the binomial count \(N_5\sim\operatorname{Binomial}(5,0.4)\). The target is met by trial five exactly when \(N_5\ge3\), so calculate the terms for \(k=3,4,5\).

For exactly three successes, \(\binom{5}{3}(0.4)^3(0.6)^2=10(0.064)(0.36)=0.2304\).

For exactly four successes, \(\binom{5}{4}(0.4)^4(0.6)=5(0.0256)(0.6)=0.0768\).

For exactly five successes, \(\binom{5}{5}(0.4)^5=1(0.01024)=0.01024\).

Adding the disjoint cases gives \(P(T_3\le5)=0.2304+0.0768+0.01024=0.31744\). The probability, rounded to four decimal places, is 0.3174.

Check. The terms include every possible success count that meets the target, from three through five. The common mistake is to use only the exactly-three term, \(0.2304\), which omits outcomes with four or five successes.

Worked Example: Two Successful Inspections by Trial Four

A hypothetical inspection catches a particular defect with probability \(p=0.3\) on each independent inspection. Let \(T_2\) be the inspection on which the second defect is caught. Find the probability that both required catches have occurred by inspection four.

Method. Here \(N_4\sim\operatorname{Binomial}(4,0.3)\), and the target is met if \(N_4\ge2\). It is efficient to subtract the cases with zero or one success from one.

The probability of zero successes is \(\binom{4}{0}(0.3)^0(0.7)^4=1(1)(0.2401)=0.2401\).

The probability of one success is \(\binom{4}{1}(0.3)(0.7)^3=4(0.3)(0.343)=1.2(0.343)=0.4116\).

Thus the probability of not reaching two successes is \(0.2401+0.4116=0.6517\), and \(P(T_2\le4)=1-0.6517=0.3483\).

As a check, the direct tail gives \(P(N_4=2)=6(0.09)(0.49)=0.2646\), \(P(N_4=3)=4(0.027)(0.7)=0.0756\), and \(P(N_4=4)=0.0081\). Their sum is \(0.2646+0.0756+0.0081=0.3483\). The two methods agree. The common mistake is to subtract only the zero-success case and forget that one success also falls short of the target.

Worked Example: Reaching Two Successes by Six Versus Exactly on Six

A hypothetical monitoring procedure produces a success on each independent trial with probability \(p=0.25\). Let \(T_2\) be the trial of the second success. Calculate both the probability of reaching two successes by trial six and the probability that the second success occurs exactly on trial six.

By the deadline. Since \(N_6\sim\operatorname{Binomial}(6,0.25)\), use the complement of zero or one success. The probability of zero successes is \((0.75)^6=0.177978515625\). The probability of one success is \(\binom{6}{1}(0.25)(0.75)^5=6(0.25)(0.2373046875)=0.35595703125\).

Therefore, \(P(T_2\le6)=1-(0.177978515625+0.35595703125)=0.466064453125\), or 0.4661 rounded to four decimal places.

Exactly on trial six. For the second success to occur on trial six, trial six must be a success and exactly one of the first five trials must be a success. The probability is \(\binom{5}{1}(0.25)^2(0.75)^4=5(0.0625)(0.31640625)=0.098876953125\), or 0.0989 rounded to four decimal places.

The exactly-on-six event is only one part of the by-six event. In particular, the latter also includes completion on trials two through five. The common mistake is to use the exactly-on-six probability as the deadline probability.

Reading the Event Correctly

A reliable first step is to state the event in words and then translate it into a count. “At least \(r\) successes in the first \(n\) trials” is the same event as “the \(r\)th success occurs no later than trial \(n\).” By contrast, “the \(r\)th success occurs on trial \(n\)” requires a success on trial \(n\) and exactly \(r-1\) successes in the first \(n-1\) trials. These are different events, so their probabilities are generally different.

The exact-time expression, when \(n\ge r\), is

$$ P(T_r=n)=\binom{n-1}{r-1}p^r(1-p)^{n-r}. $$

It follows by requiring the first \(n-1\) trials to contain \(r-1\) successes and then requiring trial \(n\) to be a success. It is useful as a distinction and a check, but a deadline probability requires all possible completion times through \(n\), or equivalently the binomial tail. Do not substitute the exact-time formula for the cumulative deadline formula.

Key takeaway: Under independent Bernoulli trials with constant success probability \(p\), reaching \(r\) successes by trial \(n\) is the binomial event \(N_n\ge r\). Sum the binomial probabilities from \(r\) to \(n\), or subtract the probabilities for zero through \(r-1\) successes from one.

Check Your Understanding

For each problem, assume trials are independent and the stated success probability remains constant. Round final probabilities to four decimal places.

  1. A hypothetical sensor flags a test condition with probability \(0.2\) on each trial. What is the probability of at least two flags in the first five trials? (Answer: 0.2627)
  2. A hypothetical quality screen passes an item with probability \(0.6\) on each independent attempt. What is the probability of at least three passes in four attempts? (Answer: 0.4752)
  3. A hypothetical signal is detected with probability \(0.35\) per independent scan. Find the probability of at least two detections in three scans. (Answer: 0.2818)
  4. A hypothetical process produces an acceptable batch with probability \(0.1\) on each independent run. Find the probability of reaching two acceptable batches by run eight, using the complement. (Answer: 0.1869)
  5. A hypothetical equipment check succeeds with probability \(0.4\) on each independent trial. Find the probability that the third success occurs by trial five but not by trial four. (Answer: 0.1382)

Previous

Next

Back to Tutorials
Clinical Biostats