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Binomial distributions · Tutorial 348 of 1000

Using binomcdf for At Most Probabilities

Use binomcdf to find and interpret the cumulative probability of getting no more than a specified number of successes.

Intermediate 9 min read

What You'll Learn

  • Translate “at most” into the event \(X \leq k\).
  • Enter the number of trials, success probability, and upper cutoff in binomcdf.
  • Explain why a cumulative probability includes every count from zero through the cutoff.
  • Check a binomcdf result by adding exact binomial probabilities.
  • Distinguish binomcdf from binompdf and interpret a result in context.

From Exact Probabilities to Cumulative Probabilities

In Using binompdf for Exact Probabilities, you learned to use binompdf to find the probability that a binomial random variable equals one particular value. But many questions ask about several possible counts at once. For example, a packaging team might want the probability that at most two items in a batch are damaged. That event includes zero, one, or two damaged items—not just exactly two.

The command binomcdf finds this kind of cumulative probability for a binomial random variable. The letters “cdf” stand for cumulative distribution function. In this setting, the command adds the probabilities from zero successes through a specified upper cutoff. This is the calculator version of the cumulative probabilities for discrete variables discussed earlier in the course.

Definition: For a binomial random variable \(X\), \(\operatorname{binomcdf}(n,p,k)\) gives \(P(X\leq k)\). Here, \(n\) is the fixed number of trials, \(p\) is the probability of success on each trial, and \(k\) is the upper cutoff. The result includes every possible count from 0 through \(k\).
$$ \operatorname{binomcdf}(n,p,k) =P(X\leq k) =\sum_{x=0}^{k}P(X=x) $$

Each term in the sum is an exact probability that could be calculated with the binomial formula or binompdf. binomcdf does the addition for you. The input order is \(n\), then \(p\), then \(k\), just as binompdf takes \(n\), \(p\), and an exact count. On a TI-84-style calculator, the command is generally available through the distribution menu; menu paths can vary by model.

Translate “At Most” Before You Calculate

As in Defining n, p, and X in Context, first identify what one trial is, what counts as success, and what \(X\) counts. Then translate the wording into an inequality. “At most \(k\)” means \(X\leq k\): the cutoff is included, and all smaller possible values are included too. For a count of successes, those values start at zero.

Input check: Enter the fixed trial count \(n\), the success probability \(p\) between 0 and 1, and the whole-number upper cutoff \(k\), in that order. For a usual at-most question, \(k\) is a whole number from 0 through \(n\).

Do not confuse the exact and cumulative commands. \(\operatorname{binompdf}(n,p,k)\) gives \(P(X=k)\), while \(\operatorname{binomcdf}(n,p,k)\) gives \(P(X\leq k)\). The same cutoff value appears in each command, but the events are different. In particular, “at most two” includes \(X=0\), \(X=1\), and \(X=2\).

As with binompdf, binomcdf evaluates a model; it does not verify that the model is appropriate. Use the BINS checklist from the earlier tutorials. The trials need binary outcomes, independence, a fixed number of trials, and the same probability of success on every trial. If any of these conditions is not reasonable, the binomial calculation may not describe the situation well.

Worked Example: At Most Two Damaged Packages

Worked Example: At Most Two Damaged Packages

A warehouse uses a model in which each package has a 0.20 probability of being damaged, independently of the other packages. Six packages are selected. Let \(X\) be the number of selected packages that are damaged. Find the probability that at most two are damaged.

State. The event is \(X\leq 2\), not just \(X=2\). The random variable \(X\) counts damaged packages among the six selected.

Plan. The model specifies two outcomes for each package (damaged or not damaged), independent trials, a fixed \(n=6\), and the same \(p=0.20\) for each package. These meet the BINS conditions as stated. Use \(\operatorname{binomcdf}(6,0.20,2)\).

Do. The calculator gives:

$$ \operatorname{binomcdf}(6,0.20,2)=0.90112\approx 0.9011 $$

Check by adding the exact probabilities for zero, one, and two damaged packages. The binomial formula gives the following terms:

$$ \begin{aligned} P(X\leq 2) &=P(X=0)+P(X=1)+P(X=2)\\ &=(0.8)^6+6(0.2)(0.8)^5+\binom{6}{2}(0.2)^2(0.8)^4\\ &=0.262144+0.393216+0.245760\\ &=0.901120\\ &\approx 0.9011. \end{aligned} $$

Conclude. According to this model, the probability that at most two of the six packages are damaged is about \(0.9011\), or 90.11%. The event includes zero, one, or two damaged packages.

The exact-probability check confirms the cumulative meaning: the result is a sum of three probabilities. Using only binompdf for \(k=2\) would give \(P(X=2)=0.2458\), which answers “exactly two,” not “at most two.”

Worked Example: At Most Four Successful Checks

Worked Example: At Most Four Successful Checks

A quality-control model assigns each independent inspection a 0.30 probability of finding a flaw. Ten items are inspected. Let \(X\) be the number of inspections that find a flaw. Find the probability that at most four inspections find one.

Identify the event. “At most four” means \(X\leq 4\), so the possible counts included are 0, 1, 2, 3, and 4.

Check the model and calculate. Each inspection has two outcomes (find a flaw or do not find one), the ten inspections are independent under the stated model, the number of trials is fixed at \(n=10\), and the success probability is constant at \(p=0.30\). Thus:

$$ \operatorname{binomcdf}(10,0.30,4)\approx 0.8497 $$

Verify by summing exact probabilities. The probabilities from zero through four are:

$$ \begin{aligned} P(X\leq 4) &=\sum_{x=0}^{4}\binom{10}{x}(0.30)^x(0.70)^{10-x}\\ &=0.0282475+0.1210608+0.2334744+0.2668279+0.2001209\\ &=0.8497315\text{ (approximately)}\\ &\approx 0.8497. \end{aligned} $$

The displayed exact probabilities are rounded, so their sum can differ slightly from the sum of the unrounded probabilities used by the calculator. Using the unrounded terms gives \(0.8497316674\), which also rounds to \(0.8497\).

Interpret in context. Under this model, there is about a 0.8497 probability, or 84.97%, that no more than four of the ten inspections find a flaw.

The upper cutoff is included: the term for \(X=4\) is part of the cumulative probability. Leaving out that term would instead calculate \(P(X\leq 3)\). Always connect the calculator’s last input to the endpoint in the inequality.

Worked Example: At Most One Response

Worked Example: At Most One Response

A community garden sends seven independent reminders. For each reminder, the model assigns a 0.20 probability of receiving a response. Let \(X\) be the number of reminders that receive a response. Find the probability of at most one response.

Set up. The event is \(X\leq 1\), which includes \(X=0\) and \(X=1\). The stated model has binary outcomes (response or no response), independent trials, a fixed \(n=7\), and the same \(p=0.20\) for each reminder.

Calculate with binomcdf.

$$ \operatorname{binomcdf}(7,0.20,1)=0.5767168\approx 0.5767 $$

Check by adding the two exact probabilities.

$$ \begin{aligned} P(X\leq 1) &=P(X=0)+P(X=1)\\ &=\binom{7}{0}(0.20)^0(0.80)^7 +\binom{7}{1}(0.20)^1(0.80)^6\\ &=0.2097152+0.3670016\\ &=0.5767168\\ &\approx 0.5767. \end{aligned} $$

Answer in context. The model gives about a 57.67% probability that zero or one of the seven reminders receives a response.

This example has only two terms in its cumulative sum, but the same command works when many counts are included. The calculator adds all the exact probabilities up to the cutoff; you do not need to enter each count separately.

Common Mistakes and AP Exam Tip

A full-credit response makes the event and the calculator command agree. Show which random variable is being counted, translate the wording into an inequality, and state what the result means in the situation.

  • Using binompdf for an at-most event. binompdf gives the probability of one exact count. For “at most four,” use binomcdf with an upper cutoff of 4.
  • Leaving out smaller counts. \(P(X\leq k)\) includes every possible value from 0 through \(k\), not just \(k\) and not just values close to \(k\).
  • Excluding the endpoint. “At most four” includes four. Write \(X\leq 4\) before entering the command to keep the endpoint clear.
  • Reversing the inputs. Enter \(n\), \(p\), and then \(k\). Do not enter the number of failures or the failure probability unless the random variable and success have been defined differently.
  • Treating a calculator result as proof of a valid model. Check BINS using the context. A correct command cannot repair a model with dependent trials or changing success probabilities.
  • Giving only a decimal. State the event and interpret the answer. For example, say “The model gives a probability of about 0.5767 that zero or one reminder receives a response,” rather than writing only “0.5767.”

As a quick reasonableness check, a cumulative probability cannot be smaller than any one exact probability it includes. Also, increasing the upper cutoff can only keep the cumulative probability the same or make it larger, because the event includes at least as many possible counts.

Key Takeaway

binomcdf finds a cumulative binomial probability. Its final input is an inclusive upper cutoff: the result adds the probabilities for zero successes, one success, and every possible count up through that cutoff.

Key takeaway: Translate “at most \(k\)” into \(X\leq k\), then calculate \(\operatorname{binomcdf}(n,p,k)\). Interpret the result as the probability of \(k\) or fewer successes in the stated binomial setting.

Check Your Understanding

For each question, identify the event represented by the cumulative command. When asked, explain which exact probabilities it includes.

  1. A binomial model has \(n=8\) and \(p=0.25\). What command finds the probability of at most three successes?
  2. In a binomial setting with \(n=9\) and \(p=0.40\), what event does \(\operatorname{binomcdf}(9,0.40,2)\) represent? List the counts included.
  3. Explain the difference between \(\operatorname{binompdf}(12,0.15,4)\) and \(\operatorname{binomcdf}(12,0.15,4)\).
  4. A calculator reports \(\operatorname{binomcdf}(7,0.20,1)=0.5767\). Interpret this result if \(X\) counts reminders that receive a response.
  5. Why must the probability for exactly four successes be included when finding the probability of at most four successes?