Geometric Cumulative Probabilities and Survival Tails

A geometric model describes repeated attempts that continue until the first success. In many applications, the question is not the probability of success on one particular attempt, but the probability that success has occurred by a deadline—or that it has not yet occurred. These are cumulative and survival probabilities. Their formulas are simple, but a one-trial indexing shift can change the answer.

For example, a hypothetical laboratory may repeat a calibration procedure until it meets a specified standard. A geometric calculation is appropriate only if each attempt has the same success probability and the attempts are independent. The formulas below rely on those assumptions; they do not automatically apply when success chances change over time or attempts are dependent.

Definition: Suppose trials are independent Bernoulli trials, each with success probability $p$, where $0<p\leq 1$. Let $q=1-p$, and let $T$ be the trial number on which the first success occurs. Then $T$ has the geometric distribution on $1,2,3,\ldots$, with $$ P(T=j)=q^{j-1}p,\qquad j=1,2,3,\ldots. $$ The cumulative probability through trial $k$ is $P(T\leq k)$. The survival probability after trial $k$ is $P(T>k)$, the probability that the first success has not occurred in the first $k$ trials.

Cumulative Probability and Survival Formulas

To calculate $P(T\leq k)$, add the probabilities that the first success occurs on trial 1, trial 2, and so on, through trial $k$. To calculate $P(T>k)$, note that the first $k$ trials must all be failures. Independence makes that probability $q^k$.

P(T\leq k)=1-q^k,\qquad P(T>k)=q^k,\qquad k=0,1,2,\ldots.

The first two quantities are complements: $P(T\leq k)+P(T>k)=1$. The boundary case $k=0$ is consistent: no trial has occurred, so $P(T\leq 0)=0$ and $P(T>0)=1$. If $p=1$, then $q=0$, and the first trial is certainly a success; the same formulas apply.

Important distinction: Under the trials-until-success convention, “after trial $k$” means $T>k$, with probability $q^k$. “On or after trial $k$” means $T\geq k$, with probability $q^{k-1}$ for $k\geq 1$. The latter requires only the first $k-1$ trials to be failures.

The relation for “on or after trial $k$” also follows by taking the complement of success by trial $k-1$:

P(T\geq k)=1-P(T\leq k-1)=1-(1-q^{k-1})=q^{k-1}.

A second support convention counts the number of failures before the first success. If $W=T-1$, then $W$ takes values $0,1,2,\ldots$. Its cumulative and survival probabilities are shifted by one index:

P(W\leq r)=1-q^{r+1},\qquad P(W>r)=q^{r+1},\qquad r=0,1,2,\ldots.

Always identify which random variable the problem uses before choosing an exponent. The trials-until-success variable starts at 1; the failures-before-success variable starts at 0.

Why the Cumulative Formula Holds

For $k\geq 1$, the event $T\leq k$ consists of the mutually exclusive cases in which the first success occurs on one of the first $k$ trials. Under the model, a first success on trial $j$ requires $j-1$ failures followed by a success, so its probability is $q^{j-1}p$. Adding those probabilities gives a finite geometric series:

\begin{aligned} P(T\leq k) &=\sum_{j=1}^{k}q^{j-1}p\\ &=p\sum_{j=0}^{k-1}q^j\\ &=p\frac{1-q^k}{1-q}\\ &=1-q^k, \end{aligned}

The third line uses the finite geometric-series formula, and the last line uses $1-q=p$. This also explains why $P(T>k)=q^k$: it is the complementary event, in which all of the first $k$ trials fail.

Identify the model and support.
Confirm that trials are independent, the success probability $p$ is constant, and whether the variable counts trials or failures.

Write the event using the variable.
For a trials-until-success variable, “by trial $k$” is $T\leq k$, while “after trial $k$” is $T>k$.

Choose the cumulative or tail formula.
Use $1-q^k$ for $P(T\leq k)$ and $q^k$ for $P(T>k)$.

Check the complement and the index.
Verify that a cumulative probability and its corresponding survival probability sum to 1, and confirm how many failures the event requires.

Worked Calculations

Worked Example: Success by a Calibration Deadline

A hypothetical instrument is calibrated repeatedly. Each attempt meets the calibration standard with probability $p=0.20$, independently of all other attempts. Let $T$ be the attempt number of the first successful calibration. Find (a) the probability calibration succeeds by attempt 3 and (b) the probability it has not succeeded after attempt 3.

Method. This is a trials-until-first-success model. The failure probability is $q=1-p=1-0.20=0.80$.

(a) “By attempt 3” is $T\leq 3$. Therefore,

P(T\leq 3)=1-q^3=1-(0.80)^3=1-0.512=0.488.

The probability is $0.4880$ to four decimal places.

(b) “Not succeeded after attempt 3” is $T>3$, so the first three attempts must all fail:

$$ P(T>3)=q^3=(0.80)^3=0.512. $$

The probability is $0.5120$ to four decimal places. As a check, $0.488+0.512=1$. The common mistake this avoids is treating “after attempt 3” as requiring only two failures; the first three attempts must fail, so the exponent is 3.

Worked Example: First Valid Transmission After a Given Trial

A hypothetical device sends repeated transmissions until one is received without error. Each transmission is received without error with probability $p=0.30$, independently. Let $T$ be the transmission number of the first valid reception. Find the probability that the first valid reception occurs after transmission 5 but no later than transmission 8.

Method. “After transmission 5” and “no later than transmission 8” together describe $5<T\leq 8$. This event is the probability of surviving beyond trial 5 minus the probability of surviving beyond trial 8.

The failure probability is $q=1-0.30=0.70$. Hence,

\begin{aligned} P(5<T\leq 8) &=P(T>5)-P(T>8)\\ &=q^5-q^8\\ &=(0.70)^5-(0.70)^8\\ &=0.16807-0.05764801\\ &=0.11042199. \end{aligned}

Rounded only at the final step, the answer is $0.1104$ to four decimal places. As a check, this is positive because surviving beyond trial 5 is more likely than surviving beyond trial 8. The subtraction removes outcomes whose first valid reception occurs after trial 8.

A common error is to use $q^4-q^7$, shifting both exponents down by one. That would apply if the event were based on a failures-before-success count rather than the trial number $T$. Here, surviving beyond trial 5 requires five failures, and surviving beyond trial 8 requires eight.

Worked Example: Cumulative Probability for a Failure Count

A hypothetical inspection process checks items independently, with probability $p=0.40$ that any one item passes. Let $W$ be the number of failed inspections before the first pass. Find the probability of at most 3 failures before the first pass, and the probability of more than 3 failures.

Method. This problem counts failures, so $W$ starts at 0. The failure probability is $q=1-0.40=0.60$. At most 3 failures means the first pass occurs among trials 1 through 4.

P(W\leq 3)=1-q^{3+1}=1-(0.60)^4=1-0.1296=0.8704.

More than 3 failures means at least four failures before the first pass:

P(W>3)=q^{3+1}=(0.60)^4=0.1296.

The answers are $0.8704$ and $0.1296$, respectively, and they sum to 1. In terms of $T=W+1$, these events are $T\leq 4$ and $T>4$. The common mistake is using $q^3$ for $P(W>3)$; the event requires four failures, not three.

Common Exam Traps and Checks

The most reliable protection against an off-by-one error is to translate the event into the random variable before substituting a formula. For a trials-until-success variable, write the boundary explicitly: $T\leq k$ includes trial $k$, while $T>k$ means the first $k$ trials have all failed. For a failures-before-success variable, at most $r$ failures means success occurs by trial $r+1$.

For a valid cumulative probability, $P(T\leq k)$ must be between 0 and 1 and must increase, or stay unchanged, as $k$ increases. The survival probability $P(T>k)$ must be between 0 and 1 and must decrease, or stay unchanged, as $k$ increases. The two probabilities at the same boundary must add to 1. These checks can catch an incorrect exponent or an accidental use of a cumulative probability in place of its tail.

Key takeaway: With $T$ counting trials until the first success and $q=1-p$, $P(T\leq k)=1-q^k$ and $P(T>k)=q^k$. If the variable counts failures before success, account for the one-trial shift before calculating.

Check Your Understanding

Assume independent trials with the stated fixed success probability in each problem. Give probabilities to four decimal places.

A hypothetical sensor registers a valid reading on each attempt with probability $0.15$. If $T$ is the attempt number of the first valid reading, find $P(T>4)$. (Answer: $0.5220$)
A hypothetical seed-treatment process succeeds on each independent batch with probability $0.30$. Find the probability its first success occurs by batch 6. (Answer: $0.8824$)
A hypothetical network link sends a usable packet with probability $0.40$ per independent transmission. Find the probability the first usable packet occurs after transmission 3 but by transmission 6. (Answer: $0.1693$)
A hypothetical quality check passes each item with probability $0.20$. Let $W$ count failures before the first pass. Find $P(W\geq 5)$. (Answer: $0.3277$)
A hypothetical data relay succeeds independently on each attempt with probability $0.25$. Let $T$ be the attempt number of the first success. Find the probability that success occurs after attempt 4 but no later than attempt 7. (Answer: $0.1829$)

Geometric Cumulative Probabilities and Survival Tails

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