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Binompdf and binomcdf functions | Random variables | AP Statistics | Khan Academy

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Reviewed learning material · Video analysis · English
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Using a free-throw count X, this complete lesson contrasts the TI-84 functions binompdf and binomcdf. The first question asks for exactly four successes and gives approximately 0.14; the second asks for fewer than five, which for an integer count means at most four, and gives approximately 0.94. Both commands use n=7n=7 and p=0.35p=0.35. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to using a graphing calculator for binomial questions0:18Defining the binomial random variable X0:58Translating the question into P(X=4)P(X = 4)1:06Introducing binompdf and entering the first two arguments1:49Introduction to the Problem2:06Using the TI-84 Calculator3:15Result and Next Steps3:38Problem Review: Exact Probability3:48New Question: Less Than 5 Successes3:58Converting Inequality to Cumulative Form4:33Calculator Demonstration (binomcdf)5:13Final Result and Conclusion

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The lesson uses a graphing calculator to distinguish a single count probability from a cumulative probability in a binomial model.

Define X as the number of made free throws. The target is four made shots out of seven, with single-shot probability 0.35; n=7n=7 and p=0.35p=0.35. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

With X defined, the event "making 4 out of 7 free throws" is rewritten as P(X=4)P(X = 4). This is an exact-count probability, so the next step is to evaluate a binomial probability mass function at x=4x = 4.

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density. The inputs are trial count n, success probability p and target count x. The board first enters 7 and 0.35 and labels their meanings.

This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

We are given a scenario where a person has a 0.35 probability of making a free throw. The goal is to find the probability of making exactly 4 out of 7 free throws. This is a classic binomial probability problem.

To solve this efficiently, we can use the binompdf function on a graphing calculator like the TI-84. First, access the distribution menu by pressing 2nd and then VARS.

Scroll down to find binompdf( and select it. The function requires three pieces of information: the number of trials, the probability of success, and the specific number of successes you want to find the probability for.

Enter 7 for trials, 0.35 for p and 4 for the target count x. The chosen event is exactly four successes.

The command binompdf(7, 0.35, 4) displays 0.1442381992, a finite decimal approximation to the probability.

Rounding this to two decimal places, we find that the probability of making exactly 4 out of 7 free throws is approximately 0.14.

The next question changes the event to fewer than five made shots; the following part of this same video develops its cumulative form.

Use the same binomial model: each free throw has success probability 0.35, with 7 trials. X counts made shots.

Previously, we calculated the probability of making exactly 4 shots using binompdf. Now, we want to find the probability of making *less than 5* shots.

Since X is a discrete variable counting whole numbers, 'less than 5' means X can be 0, 1, 2, 3, or 4. This is mathematically equivalent to saying X is less than or equal to 4.

Writing it as P(X≤4)P(X \le 4) allows us to use the binomial cumulative distribution function, or binomcdf, on our calculator. Note that the input '4' here represents the upper bound of the sum, not a single exact outcome.

On a TI-84 calculator, navigate to the DISTR menu and select binomcdf. Enter the parameters: trials = 7, p=0.35p = 0.35, and x value = 4.

The calculator returns approximately 0.9444. This confirms that the probability of making fewer than 5 free throws is about 94%, which is significantly higher than the probability of making exactly 4.

Knowledge cards

01

Binomial distribution

X counts successes in seven trials with p=0.35p=0.35. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

X∼Bin⁡(7,0.35)X\sim\operatorname{Bin}(7,0.35)
02

Exact-count probability notation

Once X counts made free throws, the question "making 4 out of 7" is expressed as P(X=4)P(X = 4). This notation asks for the probability of exactly 4 successes, not a range of outcomes.

P(X=4)P(X = 4)
03

binompdf on the TI-84

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density. The three inputs are n, p and x; the remainder of the same video completes and evaluates the command.

binompdf⁡(n,p,x)\operatorname{binompdf}(n, p, x)
04

Label calculator parameters

The red arrow identifies the trial count n in the calculator input. Clearly identifying the parameters helps explain the calculation; this is a presentation tip, a way to make the calculation easier to follow.

05

Calculating Binomial Probability with binompdf

The inputs n, p and x specify the trial count, common success probability and target success count in the binomial model.

binompdf(n,p,x)binompdf(n, p, x)
06

Example: Free Throw Probability

If a player has a 0.35 chance of making a free throw, the probability of making exactly 4 out of 7 attempts can be calculated using binompdf(7, 0.35, 4). The result is approximately 0.14.

P(X=4)=binompdf(7,0.35,4)≈0.14P(X = 4) = binompdf(7, 0.35, 4) ≈ 0.14
07

Binomial Distribution Parameters

A binomial setting requires a fixed number of independent trials (n) and a constant probability of success (p). The random variable X counts the number of successes.

X∼B(n,p)X \sim B(n, p)
08

Exact vs. Cumulative Probability

Use binompdf(n, p, x) to find P(X=x)P(X = x), the probability of exactly x successes. Use binomcdf(n, p, x) to find P(X≤x)P(X \le x), the probability of x or fewer successes.

P(X=x)→binompdf;P(X≤x)→binomcdfP(X=x) \rightarrow \text{binompdf}; \quad P(X \leq x) \rightarrow \text{binomcdf}
09

Handling Strict Inequalities

For an integer-valued count X and integer threshold k, X<kX<k is equivalent to X≤k−1X\le k-1. Here k=5k=5, so the calculator upper bound is 4.

P(X<k)=P(X≤k−1)P(X < k) = P(X \leq k-1)
10

Calculator Syntax for binomcdf

When entering data into binomcdf, the 'x value' field specifies the maximum number of successes to include in the cumulative sum. It does not calculate the probability of that specific number alone.

binomcdf⁡(n,p,k)\operatorname{binomcdf}(n,p,k)

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 13

X

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Handwritten line: "X = # of made FTs from 7 trials with p=0.35p = 0.35".

  2. Audio
    Observation

    The explanation defines X as the free-throw success count and records the common single-shot probability.

Symbol

X

Meaning

Binomial random variable representing the number of made free throws out of 7 trials.

Domain

Integer-valued count; in this example, possible values are 0 through 7.

p

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Printed problem states "I have a 0.35 probability of making a free throw." Handwritten annotation adds "p=0.35p = 0.35".

  2. Audio
    Observation

    The explanation identifies p as the single-trial success probability.

Symbol

p

Meaning

Probability of success on one trial, here the probability of making a single free throw.

Domain

Real number between 0 and 1; in this example p=0.35p = 0.35.

n

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The handwritten expression becomes "binompdf(7," and a red arrow labels the 7 as "n".

  2. Audio
    Observation

    The explanation labels the trial-count argument n with a red arrow.

Symbol

n

Meaning

Number of trials in the stated mutually independent binomial model.

Domain

Positive integer; in this example n=7n = 7.

binompdf

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Handwritten function name "binompdf" appears, followed by an opening parenthesis and arguments being entered.

  2. Audio
    Observation

    The explanation introduces binompdf and enters its first trial-count and success-probability arguments.

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Symbol

binompdf

Meaning

Graphing-calculator function for evaluating a binomial probability mass function at a specified number of successes.

Domain

Applied to a binomial model with parameters n, p, and a target success count x.

X

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Handwritten text defines X = # of made FTs from 7 trials.

Symbol

X

Meaning

The number of successful free throws (made free throws) out of the total trials.

Domain

Integer values from 0 to 7.

n

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    Handwritten text shows '7 trials' with a red arrow pointing to 'n'.

Symbol

n

Meaning

Number of trials in the stated mutually independent binomial model.

Domain

Positive integer; here n=7n = 7.

p

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Handwritten text shows p=0.35p = 0.35 with a red arrow pointing to 'p'.

Symbol

p

Meaning

The probability of success on a single trial.

Domain

Real number between 0 and 1; here p=0.35p = 0.35.

binompdf

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    Handwritten expression binompdf(7, 0.35, 4).

Symbol

binompdf

Meaning

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density.

Domain

Takes arguments (trials, p, x value).

X

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The whiteboard defines X = # of made FTs from 7 trials with p=0.35p = 0.35.

Symbol

X

Meaning

The number of successful free throws made out of 7 attempts.

Domain

Discrete random variable taking integer values from 0 to 7.

n

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The whiteboard shows '7 trials' and the calculator input uses 7 for trials.

Symbol

n

Meaning

Number of trials in the stated mutually independent binomial model.

Domain

Positive integer, specifically n=7n = 7 in this example.

p

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The whiteboard states 'p=0.35p = 0.35' and the calculator input uses 0.35 for p.

Symbol

p

Meaning

The probability of success on a single trial (making a free throw).

Domain

Real number between 0 and 1, specifically p=0.35p = 0.35.

binompdf(n, p, x)

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The whiteboard shows binompdf(7, 0.35, 4) ≈ 0.14.

Symbol

binompdf(n, p, x)

Meaning

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density.

Domain

Returns a real number between 0 and 1.

Knowledge points · 8

Binomial random variable setup for repeated free throws

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The explanation defines X as the free-throw success count and records the common single-shot probability.

  2. Formula
    Observation

    Handwritten definition: "X = # of made FTs from 7 trials with p=0.35p = 0.35".

Definition
Explanation

Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Formula
X∼Bin⁡(7,0.35)X\sim\operatorname{Bin}(7,0.35)
Conditions
  1. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Translating the word problem into P(X=4)P(X = 4)

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Defining the success count lets the explanation express the first question as the exact-count event.

  2. Formula
    Observation

    Handwritten expression: "P(X=4)P(X = 4)".

Formula
Explanation

Once X is defined as the number of made free throws, the event "making 4 out of 7 free throws" is represented by the exact-count probability P(X=4)P(X = 4). This is a point probability, not a cumulative probability.

Formula
P(X=4)P(X = 4)
Conditions
  1. X must be defined as the number of made free throws.

  2. The desired outcome is exactly 4 made free throws out of 7.

Prerequisites
  1. Binomial random variable setup for repeated free throws

TI-84 binompdf function for exact binomial probabilities

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The explanation introduces binompdf and enters its first trial-count and success-probability arguments.

  2. Formula
    Observation

    Handwritten expression develops into "binompdf(7, 0.35" with a red arrow labeling 7 as n.

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Method
Explanation

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density.

Formula
binompdf⁡(n,p,x)\operatorname{binompdf}(n, p, x)
Conditions
  1. Use when the random variable follows a binomial model.

  2. n is the number of trials.

  3. p is the probability of success on each trial.

  4. x is the exact number of successes whose probability is desired.

  5. At this early time the command is being introduced; the same full video later enters x and computes the answer.

Prerequisites
  1. Binomial random variable setup for repeated free throws
  2. Translating the word problem into P(X=4)P(X = 4)

Label calculator parameters

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The lesson recommends labeling the trial parameter and illustrates it with a red arrow.

  2. Formula
    Observation

    A red arrow points from the handwritten 7 in binompdf(7, 0.35 to the label "n".

Method
Explanation

The red arrow identifies the trial count n in the calculator input. Clearly identifying the parameters helps explain the calculation; this is a presentation tip, a way to make the calculation easier to follow.

Formula
Prerequisites
  1. TI-84 binompdf function for exact binomial probabilities

Using binompdf on a TI-84 Calculator

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The calculator demonstration enters the parameters for the exact-count event and obtains the displayed decimal approximation.

  2. Formula
    Observation

    Handwritten text shows binompdf(7, 0.35, 4).

Method
Explanation

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density. Its x input is the target count, not the expectation of X.

Formula
binompdf(n,p,x)binompdf(n, p, x)
Conditions
  1. The experiment must follow a binomial distribution (fixed number of independent trials, two possible outcomes, constant probability of success).

  2. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Binomial Distribution Setup

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board identifies the free-throw count X, trial count 7 and common success probability 0.35.

Definition
Explanation

Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Formula
X∼B(n,p)X \sim B(n, p)
Conditions
  1. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Exact Probability using binompdf

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    Whiteboard shows P(X=4)P(X = 4) and binompdf(7, 0.35, 4) ≈ 0.14

Method
Explanation

binompdf gives the point probability mass P(X=x)P(X=x) for an integer count. Despite its calculator name, this is a discrete probability mass, not a continuous probability density.

Formula
P(X=x)=binompdf(n,p,x)P(X = x) = \text{binompdf}(n, p, x)
Conditions
  1. Requires parameters n, p, and target value x

  2. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Prerequisites
  1. Binomial Distribution Setup

Cumulative Probability using binomcdf

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The explanation converts the integer threshold to an inclusive upper bound and uses the cumulative calculator function.

  2. Formula
    Observation

    Whiteboard shows P(X<5)P(X < 5) or P(X≤4)P(X \le 4) and binomcdf(7, 0.35, 4) ≈ 0.94

Method
Explanation

To find the probability of at most x successes (or fewer than x+1x+1), use the binomial cumulative distribution function (binomcdf). This calculates P(X≤x)P(X \le x).

Formula
P(X≤x)=binomcdf(n,p,x)P(X \leq x) = \text{binomcdf}(n, p, x)
Conditions
  1. Requires parameters n, p, and upper bound x

  2. Useful for inequalities like X<kX < k by converting to X≤k−1X \le k-1

  3. For the conversion X<kX<k to X≤k−1X\le k-1, X is integer-valued and k is an integer.

  4. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Prerequisites
  1. Binomial Distribution Setup
Derivations and proofs · 2

Converting the free-throw word problem into a binomial probability statement

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    Defining the success count lets the explanation express the first question as the exact-count event.

  2. Formula
    Observation

    Written sequence: "X = # of made FTs from 7 trials with p=0.35p = 0.35" followed by "P(X=4)P(X = 4)".

Intuitive argument
Steps
  1. Expression
    Make 4 out of 7 free throws\text{Make 4 out of 7 free throws}
    Explanation

    Start from the printed question asking for the probability of making 4 out of 7 free throws when each free throw has probability 0.35.

    Justification

    Printed problem statement visible throughout the clip.

    Shown in the video
  2. Expression
    X=# of made FTs from 7 trials with p=0.35X = \text{\# of made FTs from 7 trials with } p = 0.35
    Explanation

    Define a random variable X to count the number of made free throws among the 7 attempts.

    Justification

    The lecture introduces the count variable X for the printed experiment.

    Shown in the video
  3. Expression
    P(X=4)P(X = 4)
    Explanation

    Rewrite the requested event as the probability that X takes the exact value 4.

    Justification

    Direct translation from the phrase "making 4 out of 7 free throws" once X is defined as the count of made free throws.

    Shown in the video
Conclusion

Defining the count X translates the first word problem to P(X=4)P(X=4); later in this full video the calculator evaluates it. This is event translation, not a proof of the general binomial formula.

Converting Strict Inequality to Cumulative Form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation identifies fewer than five successes as the integer-count event of at most four.

  2. Formula
    Observation

    Whiteboard writes P(X<5)P(X < 5) or P(X≤4)P(X \le 4)

Intuitive argument
Steps
  1. Explanation

    Identify the condition 'less than 5'. Since X is discrete (integer counts), the values satisfying X<5X < 5 are 0, 1, 2, 3, and 4.

    Justification

    Definition of strict inequality on integers.

    Shown in the video
  2. Explanation

    Rewrite the condition as 'less than or equal to 4'. The set of values {0, 1, 2, 3, 4} is equivalent to X≤4X \le 4.

    Justification

    Equivalence of sets for discrete variables.

    Shown in the video
  3. Explanation

    Recognize that X≤4X \le 4 matches the input format for the cumulative distribution function binomcdf(n, p, 4).

    Justification

    Definition of binomcdf.

    Shown in the video
Conclusion

P(X<5)P(X < 5) is calculated as binomcdf(7, 0.35, 4).

Worked examples · 3

Probability of making exactly 4 out of 7 free throws

Approximate timing
Supplementary explanation
Evidence
  1. Caption evidence
    Observation

    The printed problem gives the single-shot probability 0.35 and asks for exactly four made shots out of seven.

  2. Formula
    Observation

    Handwritten setup: "X = # of made FTs from 7 trials with p=0.35p = 0.35", then "P(X=4)P(X = 4)", then "binompdf(7, 0.35" with 7 labeled n.

  3. Audio
    Observation

    The explanation introduces the trial count, common success probability and exact-count event, and labels the initial calculator parameters.

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Problem

Given a 0.35 probability of making a free throw, find the probability of making 4 out of 7 free throws.

Given
  1. Probability of making one free throw is 0.35.

  2. There are 7 free throw attempts.

  3. The desired outcome is exactly 4 made free throws.

  4. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Goal

Compute P(X=4)P(X = 4) for the binomial random variable X counting made free throws.

Steps
  1. Expression
    X=# of made FTs from 7 trials with p=0.35X = \text{\# of made FTs from 7 trials with } p = 0.35
    Explanation

    Define the random variable as the number of made free throws in 7 trials.

    Justification

    The lecturer defines X as the free-throw count.

    Shown in the video
  2. Expression
    P(X=4)P(X = 4)
    Explanation

    Translate the requested event into probability notation.

    Justification

    The phrase "making 4 out of 7 free throws" means X equals 4 under the chosen definition of X.

    Shown in the video
  3. Expression
    binompdf⁡(7,0.35,…)\operatorname{binompdf}(7, 0.35, \ldots)
    Explanation

    Begin entering the calculator function for the exact binomial probability, using 7 as n and 0.35 as p.

    Justification

    The explanation enters and labels the initial calculator arguments.

    Shown in the video
Answer

This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Verification

The early interval sets up the event; the later same-video calculator result verifies approximately 0.14 under the stated model.

Probability of making exactly 4 out of 7 free throws

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The calculator demonstration enters the parameters for the exact-count event and obtains the displayed decimal approximation.

  2. Formula
    Observation

    Handwritten text shows P(X=4)P(X=4) = binompdf(7, 0.35, 4) ≈ 0.14.

  3. Animation
    Observation

    Calculator screen shows the input binompdf(7, 0.35, 4) and the output 0.1442381992.

Problem

Given a 0.35 probability of making a free throw, what is the probability of making exactly 4 out of 7 free throws?

Given
  1. n=7n = 7 trials

  2. p=0.35p = 0.35 probability of success

  3. x=4x = 4 successes

  4. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Goal

Find P(X=4)P(X = 4).

Steps
  1. Explanation

    Identify the parameters of the binomial distribution: number of trials n=7n = 7, probability of success p=0.35p = 0.35, and desired number of successes x=4x = 4.

    Justification

    Extracted directly from the problem statement.

    Shown in the video
  2. Explanation

    Use the binompdf function on a calculator with the identified parameters.

    Justification

    The binompdf function calculates the exact probability of a specific number of successes in a binomial distribution.

    Shown in the video
  3. Expression
    binompdf(7,0.35,4)binompdf(7, 0.35, 4)
    Explanation

    Input the parameters into the calculator function.

    Justification

    Following the syntax binompdf(n, p, x).

    Shown in the video
  4. Expression
    0.14423819920.1442381992
    Explanation

    The calculator displays a finite decimal approximation to the probability.

    Justification

    Result of the calculation.

    Supplementary explanation
Answer

The probability of making exactly 4 out of 7 free throws is approximately 0.14.

Verification

The result is rounded to two decimal places as shown in the handwritten notes.

Free Throw Probability Problem

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    Full problem statement and solution steps visible on whiteboard and calculator screen.

Problem

Given a 0.35 probability of making a free throw, find the probability of making exactly 4 out of 7, and the probability of making less than 5 out of 7.

Given
  1. n=7n = 7 trials

  2. p=0.35p = 0.35 probability of success

  3. Target 1: Exactly 4 successes

  4. Target 2: Less than 5 successes

  5. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.

Goal

Calculate P(X=4)P(X = 4) and P(X<5)P(X < 5).

Steps
  1. Explanation

    Define the random variable X as the number of made free throws.

    Justification

    Problem setup.

    Shown in the video
  2. Explanation

    For P(X=4)P(X = 4), use the exact probability function: binompdf(7, 0.35, 4).

    Justification

    Method for exact count.

    Shown in the video
  3. Explanation

    Calculate result: binompdf(7, 0.35, 4) ≈ 0.1442.

    Justification

    Calculator output shown on screen.

    Shown in the video
  4. Explanation

    For P(X<5)P(X < 5), convert to cumulative form: P(X≤4)P(X \le 4).

    Justification

    Discrete variable property.

    Shown in the video
  5. Explanation

    Use the cumulative function: binomcdf(7, 0.35, 4).

    Justification

    Method for 'at most' or 'less than' conditions.

    Shown in the video
  6. Explanation

    Calculate result: binomcdf(7, 0.35, 4) ≈ 0.9444.

    Justification

    Calculator output shown on screen.

    Shown in the video
Answer

P(X=4)P(X = 4) ≈ 0.14; P(X<5)P(X < 5) ≈ 0.94

Verification

Results match the calculator display shown in the video.

Visual events · 5

Initial printed problem statement

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    White background with black printed text showing the free-throw probability problem and a second question about making less than 5 free throws.

  2. Audio
    Observation

    The introduction presents the printed free-throw questions and the calculator-based binomial lesson.

Objects
  1. Printed sentence about 0.35 probability of making a free throw

  2. Printed question about making 4 out of 7 free throws

  3. Printed question about making less than 5 free throws

Changes
  1. No mathematical writing yet; the screen presents the problem context.

Invariants
  1. The success probability remains 0.35.

  2. The number of trials in the first question remains 7.

Interpretation

The visual establishes the word problem that will be modeled as a binomial random variable.

Writing the binomial random variable definition

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Blue handwriting appears progressively to form "X = # of made FTs from 7 trials with p=0.35p = 0.35".

  2. Audio
    Observation

    The explanation defines X as the free-throw success count and records the common single-shot probability.

Objects
  1. Variable X

  2. Count of made free throws

  3. Number 7

  4. Parameter p=0.35p = 0.35

Changes
  1. The abstract word problem becomes a named random variable with explicit parameters.

Invariants
  1. The trials remain 7.

  2. The success probability remains 0.35.

Interpretation

This visual step formalizes the probabilistic model needed before applying a calculator function.

Entering binompdf arguments on the whiteboard

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Blue handwriting forms "binompdf(" and then fills in "7, 0.35"; a red arrow labels the 7 as "n".

  2. Audio
    Observation

    The explanation introduces binompdf and enters its first trial-count and success-probability arguments.

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Objects
  1. Function name binompdf

  2. First argument 7

  3. Second argument 0.35

  4. Red label n

Changes
  1. The probability statement P(X=4)P(X = 4) is mapped to calculator syntax.

  2. The first two parameters of the binomial model are inserted explicitly.

Invariants
  1. The model remains binomial.

  2. The target event remains exactly 4 successes, although that value is not yet entered.

Interpretation

The handwriting connects the event to calculator syntax and labels the trial parameter n.

Navigating the TI-84 Calculator for binompdf

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    A virtual TI-84 Plus CE calculator is shown on screen. The user navigates the DISTR menu, selects binompdf, inputs the values 7, 0.35, and 4, and pastes the command to get the result.

Objects
  1. Virtual TI-84 Plus CE calculator interface

  2. DISTR menu

  3. binompdf input screen

Changes
  1. Menu scrolls from normalpdf to binompdf.

  2. Values are typed into the trials, p, and x value fields.

  3. The final command binompdf(7, 0.35, 4) is pasted and executed.

Invariants
  1. The problem statement remains visible at the top of the screen.

Interpretation

Demonstrates the practical steps to compute a binomial probability using a graphing calculator.

TI-84 Calculator Demonstration

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Screen recording of TI-84 Plus CE emulator showing menu navigation and calculation.

Objects
  1. Calculator interface

  2. DISTR menu

  3. Input fields for trials, p, x value

Changes
  1. User navigates to DISTR menu

  2. Selects binomcdf

  3. Inputs 7, 0.35, 4

  4. Result 0.9443924648 appears

Invariants
  1. Problem context remains the same throughout the demo

Interpretation

Demonstrates the practical application of the binomcdf function on a standard graphing calculator used in AP Statistics.

Misconceptions · 2

Unclear calculator inputs

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The lesson recommends labeling the trial parameter and illustrates it with a red arrow.

  2. Formula
    Observation

    The handwritten 7 in binompdf(7, 0.35 is explicitly annotated with a red arrow labeled n.

Misconception

A calculator command alone always explains what its arguments mean.

Clarification

Identify n as the trial count, p as the common success probability and x as the target count.

Misinterpreting binomcdf Input

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation distinguishes the cumulative upper-bound input from a single exact outcome.

Misconception

Thinking that the x-value in binomcdf(n, p, x) refers to the probability of exactly x successes.

Clarification

The x-value in binomcdf is the upper bound of the cumulative sum, representing P(X≤x)P(X \le x), not P(X=x)P(X = x).

Concept relations · 4

Binomial random variable setup for repeated free throws → Translating the word problem into P(X=4)P(X = 4)

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Defining the success count lets the explanation express the first question as the exact-count event.

  2. Formula
    Observation

    Written progression from X definition to P(X=4)P(X = 4).

Application
Explanation

The binomial random variable definition is applied to translate the phrase "making 4 out of 7 free throws" into the exact probability statement P(X=4)P(X = 4).

Translating the word problem into P(X=4)P(X = 4) → TI-84 binompdf function for exact binomial probabilities

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation introduces binompdf and enters its first trial-count and success-probability arguments.

  2. Formula
    Observation

    The board moves from P(X=4)P(X = 4) to binompdf(7, 0.35 ...

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Application
Explanation

The exact-count probability P(X=4)P(X = 4) is the type of quantity evaluated by the binomial probability mass function, which the video implements through binompdf.

TI-84 binompdf function for exact binomial probabilities → Label calculator parameters

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The lesson recommends labeling the trial parameter and illustrates it with a red arrow.

  2. Formula
    Observation

    A red arrow labels the first argument 7 as n inside the binompdf expression.

Application
Explanation

Labeling calculator parameters makes the probability calculation easier to interpret.

Exact Probability using binompdf → Cumulative Probability using binomcdf

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation converts the integer threshold to an inclusive upper bound and uses the cumulative calculator function.

Contrast
Explanation

binompdf calculates a single point probability P(X=k)P(X=k), while binomcdf calculates a cumulative probability P(X≤k)P(X\le k) by summing point probabilities from 0 to k.

Find an answer · 7

How do you define the binomial random variable X in the free-throw example?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Handwritten definition of X as the number of made free throws from 7 trials with p=0.35p = 0.35.

Knowledge points
  1. Binomial random variable setup for repeated free throws

Why does "making 4 out of 7 free throws" become P(X=4)P(X = 4)?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Defining the success count lets the explanation express the first question as the exact-count event.

  2. Formula
    Observation

    Handwritten P(X=4)P(X = 4).

Knowledge points
  1. Binomial random variable setup for repeated free throws
  2. Translating the word problem into P(X=4)P(X = 4)

What are the first two arguments of binompdf in this example, and what do they represent?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation introduces binompdf and enters its first trial-count and success-probability arguments.

  2. Formula
    Observation

    Handwritten binompdf(7, 0.35 with 7 labeled n.

Uncertainties
  1. This initial interval introduces the parameters; the same complete video later enters the remaining argument and evaluates both requested probabilities.

Knowledge points
  1. TI-84 binompdf function for exact binomial probabilities
  2. Label calculator parameters

Why label the parameters in a calculator probability command?

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The lesson recommends labeling the trial parameter and illustrates it with a red arrow.

  2. Formula
    Observation

    Red arrow labeling the 7 as n.

Knowledge points
  1. Label calculator parameters

How do I calculate the exact probability of a binomial distribution using a TI-84 calculator?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The calculator demonstration enters the parameters for the exact-count event and obtains the displayed decimal approximation.

Knowledge points
  1. Using binompdf on a TI-84 Calculator

How do I calculate the probability of getting less than k successes in a binomial distribution?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation converts the integer threshold to an inclusive upper bound and uses the cumulative calculator function.

Knowledge points
  1. Cumulative Probability using binomcdf
  2. Converting Strict Inequality to Cumulative Form

What is the difference between the x-value in binompdf and binomcdf?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The explanation converts the integer threshold to an inclusive upper bound and uses the cumulative calculator function.

Knowledge points
  1. Exact Probability using binompdf
  2. Cumulative Probability using binomcdf
  3. Misinterpreting binomcdf Input
Coverage and review notes

Covered · Introductory narration explains that the video will use a graphing calculator for binomial random variable questions; printed problem is visible.

Covered · The speaker reads the first question and defines X as the number of made free throws from 7 trials with p=0.35p = 0.35.

Covered · The word problem is translated into P(X=4)P(X = 4).

Covered · The function and its first two parameters are introduced and labeled; the following section completes the same calculation.

Covered · The entire clip focuses on setting up and calculating a specific binomial probability using the binompdf function on a calculator.

Covered · Initial problem setup and review of previous part (binompdf).

Covered · Explanation of the new question (less than 5) and conversion to cumulative form.

Covered · The calculator demonstration evaluates the cumulative probability and the final board retains both rounded answers.

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  • Binomial distribution ExplanationAt 0:31
    Why this connection?

    X counts successes in seven trials with p=0.35p=0.35. Model assumption: seven mutually independent Bernoulli trials, each with the same success probability 0.35. Equal single-shot success probabilities alone do not guarantee independence.