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Binomial distributions · Tutorial 349 of 1000

Finding At Least Probabilities with binomcdf

Use the complement and the cutoff one below the target to calculate “at least” binomial probabilities without an off-by-one error.

Intermediate 9 min read

What You'll Learn

  • Translate “at least k” into the event \(X\geq k\).
  • Use the complement \(X\leq k-1\) to set up the calculation.
  • Enter the correct cutoff in \(1-\operatorname{binomcdf}(n,p,k-1)\).
  • Distinguish “at least” from “more than” at an endpoint.
  • Check a binomial model with the BINS conditions and interpret the result in context.

From “At Most” to “At Least”

In Using binomcdf for At Most Probabilities, you learned that \(\operatorname{binomcdf}(n,p,k)\) gives the probability of \(k\) or fewer successes. This makes it useful for “at most” questions. For “at least” questions, we can use the same cumulative command by finding the complement of the event we want.

The key is to translate the wording before entering numbers. “At least four” includes four and every larger count, so it means \(X\geq 4\). Its complement is not \(X\leq 4\); it is \(X<4\). Because a binomial count can only be a whole number, \(X<4\) means \(X\leq 3\). That is why the cumulative command uses the cutoff one less than the “at least” threshold.

Formula: If \(X\) is a binomial random variable and \(k\) is a whole-number threshold from 1 through \(n\), then “at least \(k\)” means \(X\geq k\). Its complement is \(X\leq k-1\), so:
$$ P(X\geq k)=1-P(X\leq k-1) =1-\operatorname{binomcdf}(n,p,k-1) $$

The inputs to \(\operatorname{binomcdf}\) are still \(n\), \(p\), and an inclusive upper cutoff. The subtraction from 1 changes the event from the counts below the threshold to the counts at or above it. The final input is \(k-1\), not \(k\).

Find the Complement’s Cutoff First

A reliable method is to write the desired event, write its complement, and only then choose the cumulative cutoff. For example, “at least five” becomes \(X\geq 5\). The complement consists of zero, one, two, three, or four successes: \(X\leq 4\). Thus the calculation is \(1-\operatorname{binomcdf}(n,p,4)\).

This one-less cutoff follows from the meaning of “at least,” not from a special calculator rule. The complement must contain every possible count that does not satisfy the request. For “at least five,” five is included in the requested event, so it must not be included in the complement.

1
Write the requested event.
Translate “at least \(k\)” into \(X\geq k\).
2
Write its complement.
The complement is \(X<k\), which for a whole-number count is \(X\leq k-1\).
3
Use binomcdf and subtract.
Calculate \(1-\operatorname{binomcdf}(n,p,k-1)\).
4
Interpret the result.
Describe the probability of \(k\) or more successes in the context.

As in Defining n, p, and X in Context, identify what counts as a success and what \(X\) counts before calculating. Also use the BINS checklist from the earlier tutorials: Binary outcomes, Independent trials, a fixed Number of trials, and the Same probability of success. A calculator evaluates the model you enter; it does not establish that the model fits the situation.

Worked Example: At Least Two Successful Tests

Worked Example: At Least Two Successful Tests

A lab uses a model in which each test has a 0.15 probability of a positive result, independently of the other tests. Eight tests are run. Let \(X\) be the number of tests with a positive result. Find the probability of at least two positive results.

State. The requested event is \(X\geq 2\). Its complement is \(X<2\), or \(X\leq 1\), because \(X\) counts a whole number of positive results.

Plan. The model specifies two outcomes per test (positive or not positive), independent tests, a fixed number of trials \(n=8\), and the same probability \(p=0.15\) of a positive result on every test. The BINS conditions are met as stated. The complement’s inclusive upper cutoff is \(2-1=1\), so use \(1-\operatorname{binomcdf}(8,0.15,1)\).

Do. The calculator gives:

$$ \begin{aligned} P(X\geq 2) &=1-\operatorname{binomcdf}(8,0.15,1)\\ &=1-0.6571830310\\ &=0.3428169690\\ &\approx 0.3428. \end{aligned} $$

The cumulative value is rounded here for display; using more decimal places, \(P(X\leq 1)\approx 0.657183\), so its complement is approximately \(0.342817\). Both give \(0.3428\) to four decimal places.

Check the complement by adding its two exact probabilities:

$$ \begin{aligned} P(X\leq 1) &=P(X=0)+P(X=1)\\ &=(0.85)^8+8(0.15)(0.85)^7\\ &=0.2724905250+0.3846925059\\ &\approx 0.6571830310. \end{aligned} $$

Conclude. According to this model, the probability that at least two of the eight tests have a positive result is about \(0.3428\), or 34.28%.

Notice the endpoint: \(X=2\) belongs to “at least two,” so it must not appear in the complement. Using \(\operatorname{binomcdf}(8,0.15,2)\) as the complement would include \(X=2\) and produce the wrong event.

Worked Example: At Least Three Returned Forms

Worked Example: At Least Three Returned Forms

A school office models each mailed form as having a 0.20 probability of being returned, independently of the other forms. Ten forms are mailed. Let \(X\) be the number returned. Find the probability that at least three forms are returned.

Translate. “At least three” means \(X\geq 3\). The complement is \(X<3\), or \(X\leq 2\), so the calculator cutoff is \(3-1=2\).

Check BINS and calculate. Each form has two outcomes (returned or not returned), the model says the outcomes are independent, the number of trials is fixed at \(n=10\), and the success probability is constant at \(p=0.20\). Thus the binomial calculation is appropriate under the stated model:

$$ \begin{aligned} P(X\geq 3) &=1-\operatorname{binomcdf}(10,0.20,2)\\ &=1-0.6777995264\\ &=0.3222004736\\ &\approx 0.3222. \end{aligned} $$

The complement can also be checked by summing the probabilities for zero, one, and two returned forms:

$$ \begin{aligned} P(X\leq 2) &=(0.80)^{10} +10(0.20)(0.80)^9 +\binom{10}{2}(0.20)^2(0.80)^8\\ &=0.1073741824+0.2684354560+0.3019898880\\ &=0.6777995264. \end{aligned} $$

Interpret. Under this model, there is about a \(0.3222\) probability, or 32.22%, that at least three of the ten mailed forms are returned.

Here, the complement includes exactly the counts that fall short of three. The count of two is included in the complement, while the count of three is not. Checking those boundary values is a quick way to catch a cutoff error.

Worked Example: At Least One Successful Connection

Worked Example: At Least One Successful Connection

A technician’s model assigns each of six independent connection attempts a 0.12 probability of succeeding. Let \(X\) be the number of successful connections. Find the probability of at least one success.

Identify the event and conditions. The request is \(X\geq 1\); its complement is \(X<1\), which is \(X\leq 0\). Each attempt has two outcomes (success or failure), the attempts are independent, there are a fixed \(n=6\) attempts, and the success probability stays at \(p=0.12\). All four BINS conditions are satisfied under the stated model.

Calculate. Since the complementary event has an upper cutoff of \(1-1=0\), use:

$$ \begin{aligned} P(X\geq 1) &=1-\operatorname{binomcdf}(6,0.12,0)\\ &=1-0.4644040868\\ &=0.5355959132\\ &\approx 0.5356. \end{aligned} $$

For this particular complement, \(X\leq 0\) means exactly zero successes. Its probability is \(P(X=0)=(0.88)^6=0.464404086784\), which confirms the cumulative result. Therefore the model gives about a 53.56% probability of at least one successful connection in six attempts.

“At least one” is a useful special case of the same method. The complement is zero successes, but the method also works when the threshold is larger: subtract the probability of every count below that threshold.

Cutoff Checks and Common Mistakes

The most common error is putting the requested threshold itself into the cumulative command. If the question asks for at least \(k\), then \(\operatorname{binomcdf}(n,p,k)\) includes \(k\) in the cumulative probability. Subtracting that value from 1 gives \(P(X\geq k+1)\), not \(P(X\geq k)\). The correct cutoff is \(k-1\).

It also helps to distinguish “at least” from “more than.” “At least four” means \(X\geq 4\), so use \(1-\operatorname{binomcdf}(n,p,3)\). “More than four” means \(X>4\), or \(X\geq 5\), so use \(1-\operatorname{binomcdf}(n,p,4)\). The difference is whether the endpoint four is included in the requested event.

  • Forgetting the complement. \(\operatorname{binomcdf}\) directly gives a probability of being at or below a cutoff. Subtract that cumulative probability from 1 to get the upper tail.
  • Using \(k\) instead of \(k-1\). For “at least \(k\),” the complement stops at \(k-1\). Using \(k\) would remove the probability of exactly \(k\) from the answer.
  • Confusing “at least” and “more than.” “At least \(k\)” includes \(k\); “more than \(k\)” does not. Translate each phrase into an inequality first.
  • Entering the wrong values. Use the number of trials \(n\), the probability of the outcome defined as success \(p\), and then the complement cutoff \(k-1\), in that order.
  • Giving only a calculator result. Identify \(X\), state the event, and interpret the probability in context. A full-credit response connects the number to the situation.

For a threshold \(k=0\), “at least zero” is certain, so its probability is 1; the usual formula would require a cutoff of \(-1\), which is unnecessary. For a threshold greater than \(n\), “at least \(k\)” is impossible and has probability 0. In ordinary calculator questions, \(1\leq k\leq n\), so \(k-1\) is a valid cumulative cutoff.

AP Exam Tip: Show the event and complement before the calculator command. For example: “At least five means \(X\geq 5\); the complement is \(X\leq 4\). Therefore \(P(X\geq 5)=1-\operatorname{binomcdf}(n,p,4)\).” Then interpret the result in context.

Key Takeaway

For an “at least” probability, use the complement: find the probability of all the smaller counts, then subtract it from 1. The endpoint must be handled carefully because the requested event includes its threshold.

Key takeaway: Translate “at least \(k\)” into \(X\geq k\), then calculate \(1-\operatorname{binomcdf}(n,p,k-1)\). The final input is one less than the requested threshold because the complement ends immediately below it.

Check Your Understanding

For each question, write the event or command before calculating. Pay particular attention to the complement’s cutoff.

  1. A binomial random variable has \(n=9\) and \(p=0.30\). What command finds the probability of at least four successes?
  2. What event does \(1-\operatorname{binomcdf}(12,0.25,5)\) represent? State its inequality in terms of \(X\).
  3. Explain why \(1-\operatorname{binomcdf}(10,0.20,3)\) does not find the probability of at least three successes.
  4. Write the complement of “more than two successes” and give the appropriate complement command for \(n=8\) and \(p=0.40\).
  5. A model gives \(1-\operatorname{binomcdf}(6,0.12,0)=0.5356\). Interpret this probability if \(X\) counts successful connection attempts.

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