4.4 Binomial distribution

Achievable AP Statistics

4. Probability and random variables

Our AP Statistics course is currently in development and is a work-in-progress.

Binomial distribution

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Binomial Distribution

The binomial distribution is used when you are repeating the same random process multiple times and want to track how often a specific outcome (called a “success”) occurs. Instead of calculating each probability from scratch, this distribution gives you a structured way to model and solve these types of problems efficiently.

In this section, you’ll learn:

What a binomial distribution is and how it works
The conditions required to use it (BINS)
How to calculate probabilities using the formula
How to interpret mean, variance, and standard deviation
How to use your calculator to quickly find probabilities

As you move through the page, focus on recognizing when a situation fits a binomial setup. Once you can identify that, the rest becomes much more straightforward. :contentReference[oaicite:0]

Definitions

Binomial probability distribution: A probability distribution used when you repeat the same chance process a fixed number of times, and each trial is independent.

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Where:

$k =$ specific number of successes
$n =$ number of trials
$p =$ probability of one success Mean (expected value) for the binomial distribution : The mean is given by:
$μ = E (X) = n \cdot p$

Variance equation for the binomial distribution: The variance is given by:
$σ^{2} = Var (X) = n \cdot p \cdot (1 - p)$
Standard deviation equation for the binomial distribution: The standard deviation is given by:
$σ = n \cdot p \cdot (1 - p)$

Conditions for Using the Binomial Distribution

Sidenote

Important to know!

To use the binomial distribution, all of the “BINS” conditions must be met.

“B” stands for binary outcomes. Each trial has exactly two outcomes: “success” or “failure.”
- Example: “Flipping heads is a success; flipping tails is a failure.”
- Example: “Rolling at least an $8$ on two dice is a success; rolling anything else is a failure.”
“I” stands for independent trials. The outcome of one trial does not affect the outcome of any other trial.
“N” stands for number of trials is fixed. You decide the number of trials in advance.
“S” stands for same probability of success. The probability of success $p$ is the same on every trial.

Practice problems (with calculator, but without software):

Example 1:

A baseball player’s probability of hitting a home run is $20%$ . What is the probability that the baseball player will hit a home run three times in a row?

Solution:

(spoiler)

The probability of one home run is $20% = 0.2$ . Because the trials are independent, multiply the probabilities:

$0.2 * 0.2 * 0.2 = 0. 2^{3} = 0.008$

Example 2:

What is the probability that the baseball player will not hit any home runs in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, the player bats $10$ times, so $n = 10$ . The probability of a home run is $p = 0.2$ . “No home runs” means $k = 0$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 0) = (0 10) 0. 2^{0} (1 - 0.2)^{10 - 0} = 1 * 1 * (0.8)^{10} = 0.1074$

So, the probability of no home runs in the next 10 at-bats is about $10.74%$ .

Example 3:

What is the probability that the baseball player will hit exactly one home run in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, $n = 10$ , $p = 0.2$ , and “exactly one home run” means $k = 1$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 1) = (1 10) 0. 2^{1} (1 - 0.2)^{10 - 1} = 10 * 0.2 * (0.8)^{9} = 0.2684$

So, the probability of exactly one home run in the next 10 at-bats is about $26.84%$ .

Example 4:

What is the probability that the baseball player will hit more than one home run in his next 100 times up to bat?

Solution:

(spoiler)

To find $P (X > 1)$ directly, you’d need to add $P (X = 2) + P (X = 3) + \dots + P (X = 100)$ , which is not practical by hand. Instead, use the complement:

The opposite of “more than one” is “one or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ . From earlier parts, we have:
$P (X = 0) \approx 0.1074$
$P (X = 1) \approx 0.2684$ So,

$1 - (0.1074 + 0.2684) = 1 - 0.3758 = 0.6242$

Calculator Commands

When working with binomial probability problems, your calculator can quickly handle calculations that would otherwise take a lot of time by hand. The commands below help you find exact probabilities and cumulative probabilities for different outcomes, so it’s important to know when to use each one.

binompdf $(n, p, x)$

Gives the probability of exactly $x$ successes in $n$ trials, where $p$ is the probability of success on one trial.

Example: binompdf $(3, 0.75, 2)$ gives the probability of exactly $2$ successes in $3$ trials when the probability of success on a single trial is $0.75$ .

binomcdf $(n, p, x)$

Gives the cumulative probability of $x$ or fewer successes in $n$ trials, where $p$ is the probability of success on a single trial.

Example: binomcdf $(3, 0.75, 2)$ gives the probability of $2$ or fewer successes (that is, $0$ , $1$ , or $2$ successes) in $3$ trials when the probability of success on a single trial is $0.75$ .

Practice problems (Part B and C with software):

Example 1:

A recent survey suggested that $85%$ of new cars purchased last year were gasoline cars and the rest were electric. If $100$ new cars that were purchased last year are chosen at random, calculate the mean and standard deviation and interpret them.

Solution:

(spoiler)

$E (X) = n * p = 100 * 0.85 = 85$

Interpretation: If you repeatedly took random samples of $100$ cars purchased last year, the average number of gasoline cars per sample would be about $85$ .

$σ = n * p * (1 - p) = 100 * 0.85 * (1 - 0.85) = 12.75 = 3.5707$

Interpretation: The count of gasoline cars in samples of size $100$ will typically differ from the mean of $85$ by about $3.5707$ cars.

Example 2:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that exactly $85$ of the cars were gasoline cars?

Solution:

(spoiler)

Use binompdf $(n, p, x)$ for “exactly $x$ successes.” Here:

$n = 100$
$p = 0.85$
$x = 85$ So we compute binompdf $(100, 0.85, 85)$ .

$binompdf (100, 0.85, 85) = 0.0968$

So, there is about a $9.68%$ chance of selecting exactly $85$ gasoline cars out of $100$ .

Example 3:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that no more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“No more than $85$ ” means $85$ or fewer, so use binomcdf $(n, p, x)$ with $x = 85$ . Here:

$n = 100$
$p = 0.85$
$x = 85$

$binomcdf (100, 0.85, 85) = 0.7151$

So, there is about a $71.51%$ chance that the sample contains $85$ or fewer gasoline cars.

Example 4:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“More than $85$ ” is the complement of “ $85$ or fewer.” From the previous part:

$binomcdf (100, 0.85, 85) = 0.7151$

So,

$1 - binomcdf (100, 0.85, 85) = 1 - 0.7151 = 0.2849$

So, the probability of selecting more than $85$ gasoline cars is $0.2849$ .

Binomial probability distribution

Used for fixed number of independent, identical trials with two outcomes (success/failure)
Probability formula: $P (X = k) = (k n) p^{k} (1 - p)^{n - k}$
Key parameters:
- $n$ : number of trials
- $p$ : probability of success
- $k$ : number of successes

BINS conditions

Binary: each trial has two outcomes (success/failure)
Independent: trials do not affect each other
Number: fixed number of trials ( $n$ set in advance)
Same probability: $p$ is constant for each trial

Binomial distribution formulas

Mean (expected value): $μ = E (X) = n \cdot p$
Variance: $σ^{2} = n \cdot p \cdot (1 - p)$
Standard deviation: $σ = n \cdot p \cdot (1 - p)$

Calculator functions

binompdf $(n, p, x)$ : probability of exactly $x$ successes in $n$ trials
binomcdf $(n, p, x)$ : probability of $x$ or fewer successes (cumulative)

Problem-solving strategies

For “more than” or “at least” probabilities, use complements:
- $P (X > k) = 1 - P (X \leq k)$
Use binompdf for exact counts; binomcdf for cumulative probabilities

Interpretation of mean and standard deviation

Mean: expected number of successes in repeated samples
Standard deviation: typical deviation from the mean in sample counts

Next | 4.5 Geometric distribution

Binomial distribution

Binomial Distribution

In this section, you’ll learn:

What a binomial distribution is and how it works
The conditions required to use it (BINS)
How to calculate probabilities using the formula
How to interpret mean, variance, and standard deviation
How to use your calculator to quickly find probabilities

As you move through the page, focus on recognizing when a situation fits a binomial setup. Once you can identify that, the rest becomes much more straightforward. :contentReference[oaicite:0]{index=0}

Definitions

Binomial probability distribution: A probability distribution used when you repeat the same chance process a fixed number of times, and each trial is independent.

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Where:

$k =$ specific number of successes
$n =$ number of trials
$p =$ probability of one success Mean (expected value) for the binomial distribution : The mean is given by:
$μ = E (X) = n \cdot p$

Variance equation for the binomial distribution: The variance is given by:
$σ^{2} = Var (X) = n \cdot p \cdot (1 - p)$
Standard deviation equation for the binomial distribution: The standard deviation is given by:
$σ = n \cdot p \cdot (1 - p)$

Conditions for Using the Binomial Distribution

Sidenote

Important to know!

To use the binomial distribution, all of the “BINS” conditions must be met.

“B” stands for binary outcomes. Each trial has exactly two outcomes: “success” or “failure.”
- Example: “Flipping heads is a success; flipping tails is a failure.”
- Example: “Rolling at least an $8$ on two dice is a success; rolling anything else is a failure.”
“I” stands for independent trials. The outcome of one trial does not affect the outcome of any other trial.
“N” stands for number of trials is fixed. You decide the number of trials in advance.
“S” stands for same probability of success. The probability of success $p$ is the same on every trial.

Practice problems (with calculator, but without software):

Example 1:

A baseball player’s probability of hitting a home run is $20%$ . What is the probability that the baseball player will hit a home run three times in a row?

Solution:

(spoiler)

The probability of one home run is $20% = 0.2$ . Because the trials are independent, multiply the probabilities:

$0.2 * 0.2 * 0.2 = 0. 2^{3} = 0.008$

Example 2:

What is the probability that the baseball player will not hit any home runs in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, the player bats $10$ times, so $n = 10$ . The probability of a home run is $p = 0.2$ . “No home runs” means $k = 0$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 0) = (0 10) 0. 2^{0} (1 - 0.2)^{10 - 0} = 1 * 1 * (0.8)^{10} = 0.1074$

So, the probability of no home runs in the next 10 at-bats is about $10.74%$ .

Example 3:

What is the probability that the baseball player will hit exactly one home run in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, $n = 10$ , $p = 0.2$ , and “exactly one home run” means $k = 1$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 1) = (1 10) 0. 2^{1} (1 - 0.2)^{10 - 1} = 10 * 0.2 * (0.8)^{9} = 0.2684$

So, the probability of exactly one home run in the next 10 at-bats is about $26.84%$ .

Example 4:

What is the probability that the baseball player will hit more than one home run in his next 100 times up to bat?

Solution:

(spoiler)

To find $P (X > 1)$ directly, you’d need to add $P (X = 2) + P (X = 3) + \dots + P (X = 100)$ , which is not practical by hand. Instead, use the complement:

The opposite of “more than one” is “one or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ . From earlier parts, we have:
$P (X = 0) \approx 0.1074$
$P (X = 1) \approx 0.2684$ So,

$1 - (0.1074 + 0.2684) = 1 - 0.3758 = 0.6242$

Calculator Commands

binompdf $(n, p, x)$

Gives the probability of exactly $x$ successes in $n$ trials, where $p$ is the probability of success on one trial.

Example: binompdf $(3, 0.75, 2)$ gives the probability of exactly $2$ successes in $3$ trials when the probability of success on a single trial is $0.75$ .

binomcdf $(n, p, x)$

Gives the cumulative probability of $x$ or fewer successes in $n$ trials, where $p$ is the probability of success on a single trial.

Example: binomcdf $(3, 0.75, 2)$ gives the probability of $2$ or fewer successes (that is, $0$ , $1$ , or $2$ successes) in $3$ trials when the probability of success on a single trial is $0.75$ .

Practice problems (Part B and C with software):

Example 1:

A recent survey suggested that $85%$ of new cars purchased last year were gasoline cars and the rest were electric. If $100$ new cars that were purchased last year are chosen at random, calculate the mean and standard deviation and interpret them.

Solution:

(spoiler)

$E (X) = n * p = 100 * 0.85 = 85$

Interpretation: If you repeatedly took random samples of $100$ cars purchased last year, the average number of gasoline cars per sample would be about $85$ .

$σ = n * p * (1 - p) = 100 * 0.85 * (1 - 0.85) = 12.75 = 3.5707$

Interpretation: The count of gasoline cars in samples of size $100$ will typically differ from the mean of $85$ by about $3.5707$ cars.

Example 2:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that exactly $85$ of the cars were gasoline cars?

Solution:

(spoiler)

Use binompdf $(n, p, x)$ for “exactly $x$ successes.” Here:

$n = 100$
$p = 0.85$
$x = 85$ So we compute binompdf $(100, 0.85, 85)$ .

$binompdf (100, 0.85, 85) = 0.0968$

So, there is about a $9.68%$ chance of selecting exactly $85$ gasoline cars out of $100$ .

Example 3:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that no more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“No more than $85$ ” means $85$ or fewer, so use binomcdf $(n, p, x)$ with $x = 85$ . Here:

$n = 100$
$p = 0.85$
$x = 85$

$binomcdf (100, 0.85, 85) = 0.7151$

So, there is about a $71.51%$ chance that the sample contains $85$ or fewer gasoline cars.

Example 4:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“More than $85$ ” is the complement of “ $85$ or fewer.” From the previous part:

$binomcdf (100, 0.85, 85) = 0.7151$

So,

$1 - binomcdf (100, 0.85, 85) = 1 - 0.7151 = 0.2849$

So, the probability of selecting more than $85$ gasoline cars is $0.2849$ .

Achievable AP Statistics

4. Probability and random variables

Our AP Statistics course is currently in development and is a work-in-progress.

Binomial distribution

7 min read

Font

Discuss

Feedback

Binomial Distribution

In this section, you’ll learn:

What a binomial distribution is and how it works
The conditions required to use it (BINS)
How to calculate probabilities using the formula
How to interpret mean, variance, and standard deviation
How to use your calculator to quickly find probabilities

As you move through the page, focus on recognizing when a situation fits a binomial setup. Once you can identify that, the rest becomes much more straightforward. :contentReference[oaicite:0]

Definitions

Binomial probability distribution: A probability distribution used when you repeat the same chance process a fixed number of times, and each trial is independent.

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Where:

$k =$ specific number of successes
$n =$ number of trials
$p =$ probability of one success Mean (expected value) for the binomial distribution : The mean is given by:
$μ = E (X) = n \cdot p$

Variance equation for the binomial distribution: The variance is given by:
$σ^{2} = Var (X) = n \cdot p \cdot (1 - p)$
Standard deviation equation for the binomial distribution: The standard deviation is given by:
$σ = n \cdot p \cdot (1 - p)$

Conditions for Using the Binomial Distribution

Sidenote

Important to know!

To use the binomial distribution, all of the “BINS” conditions must be met.

“B” stands for binary outcomes. Each trial has exactly two outcomes: “success” or “failure.”
- Example: “Flipping heads is a success; flipping tails is a failure.”
- Example: “Rolling at least an $8$ on two dice is a success; rolling anything else is a failure.”
“I” stands for independent trials. The outcome of one trial does not affect the outcome of any other trial.
“N” stands for number of trials is fixed. You decide the number of trials in advance.
“S” stands for same probability of success. The probability of success $p$ is the same on every trial.

Practice problems (with calculator, but without software):

Example 1:

A baseball player’s probability of hitting a home run is $20%$ . What is the probability that the baseball player will hit a home run three times in a row?

Solution:

(spoiler)

The probability of one home run is $20% = 0.2$ . Because the trials are independent, multiply the probabilities:

$0.2 * 0.2 * 0.2 = 0. 2^{3} = 0.008$

Example 2:

What is the probability that the baseball player will not hit any home runs in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, the player bats $10$ times, so $n = 10$ . The probability of a home run is $p = 0.2$ . “No home runs” means $k = 0$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 0) = (0 10) 0. 2^{0} (1 - 0.2)^{10 - 0} = 1 * 1 * (0.8)^{10} = 0.1074$

So, the probability of no home runs in the next 10 at-bats is about $10.74%$ .

Example 3:

What is the probability that the baseball player will hit exactly one home run in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, $n = 10$ , $p = 0.2$ , and “exactly one home run” means $k = 1$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 1) = (1 10) 0. 2^{1} (1 - 0.2)^{10 - 1} = 10 * 0.2 * (0.8)^{9} = 0.2684$

So, the probability of exactly one home run in the next 10 at-bats is about $26.84%$ .

Example 4:

What is the probability that the baseball player will hit more than one home run in his next 100 times up to bat?

Solution:

(spoiler)

To find $P (X > 1)$ directly, you’d need to add $P (X = 2) + P (X = 3) + \dots + P (X = 100)$ , which is not practical by hand. Instead, use the complement:

The opposite of “more than one” is “one or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ . From earlier parts, we have:
$P (X = 0) \approx 0.1074$
$P (X = 1) \approx 0.2684$ So,

$1 - (0.1074 + 0.2684) = 1 - 0.3758 = 0.6242$

Calculator Commands

binompdf $(n, p, x)$

Gives the probability of exactly $x$ successes in $n$ trials, where $p$ is the probability of success on one trial.

Example: binompdf $(3, 0.75, 2)$ gives the probability of exactly $2$ successes in $3$ trials when the probability of success on a single trial is $0.75$ .

binomcdf $(n, p, x)$

Gives the cumulative probability of $x$ or fewer successes in $n$ trials, where $p$ is the probability of success on a single trial.

Example: binomcdf $(3, 0.75, 2)$ gives the probability of $2$ or fewer successes (that is, $0$ , $1$ , or $2$ successes) in $3$ trials when the probability of success on a single trial is $0.75$ .

Practice problems (Part B and C with software):

Example 1:

A recent survey suggested that $85%$ of new cars purchased last year were gasoline cars and the rest were electric. If $100$ new cars that were purchased last year are chosen at random, calculate the mean and standard deviation and interpret them.

Solution:

(spoiler)

$E (X) = n * p = 100 * 0.85 = 85$

Interpretation: If you repeatedly took random samples of $100$ cars purchased last year, the average number of gasoline cars per sample would be about $85$ .

$σ = n * p * (1 - p) = 100 * 0.85 * (1 - 0.85) = 12.75 = 3.5707$

Interpretation: The count of gasoline cars in samples of size $100$ will typically differ from the mean of $85$ by about $3.5707$ cars.

Example 2:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that exactly $85$ of the cars were gasoline cars?

Solution:

(spoiler)

Use binompdf $(n, p, x)$ for “exactly $x$ successes.” Here:

$n = 100$
$p = 0.85$
$x = 85$ So we compute binompdf $(100, 0.85, 85)$ .

$binompdf (100, 0.85, 85) = 0.0968$

So, there is about a $9.68%$ chance of selecting exactly $85$ gasoline cars out of $100$ .

Example 3:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that no more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“No more than $85$ ” means $85$ or fewer, so use binomcdf $(n, p, x)$ with $x = 85$ . Here:

$n = 100$
$p = 0.85$
$x = 85$

$binomcdf (100, 0.85, 85) = 0.7151$

So, there is about a $71.51%$ chance that the sample contains $85$ or fewer gasoline cars.

Example 4:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“More than $85$ ” is the complement of “ $85$ or fewer.” From the previous part:

$binomcdf (100, 0.85, 85) = 0.7151$

So,

$1 - binomcdf (100, 0.85, 85) = 1 - 0.7151 = 0.2849$

So, the probability of selecting more than $85$ gasoline cars is $0.2849$ .

Binomial probability distribution

Used for fixed number of independent, identical trials with two outcomes (success/failure)
Probability formula: $P (X = k) = (k n) p^{k} (1 - p)^{n - k}$
Key parameters:
- $n$ : number of trials
- $p$ : probability of success
- $k$ : number of successes

BINS conditions

Binary: each trial has two outcomes (success/failure)
Independent: trials do not affect each other
Number: fixed number of trials ( $n$ set in advance)
Same probability: $p$ is constant for each trial

Binomial distribution formulas

Mean (expected value): $μ = E (X) = n \cdot p$
Variance: $σ^{2} = n \cdot p \cdot (1 - p)$
Standard deviation: $σ = n \cdot p \cdot (1 - p)$

Calculator functions

binompdf $(n, p, x)$ : probability of exactly $x$ successes in $n$ trials
binomcdf $(n, p, x)$ : probability of $x$ or fewer successes (cumulative)

Problem-solving strategies

For “more than” or “at least” probabilities, use complements:
- $P (X > k) = 1 - P (X \leq k)$
Use binompdf for exact counts; binomcdf for cumulative probabilities

Interpretation of mean and standard deviation

Mean: expected number of successes in repeated samples
Standard deviation: typical deviation from the mean in sample counts

Next | 4.5 Geometric distribution

Binomial distribution

Binomial Distribution

In this section, you’ll learn:

What a binomial distribution is and how it works
The conditions required to use it (BINS)
How to calculate probabilities using the formula
How to interpret mean, variance, and standard deviation
How to use your calculator to quickly find probabilities

Definitions

Binomial probability distribution: A probability distribution used when you repeat the same chance process a fixed number of times, and each trial is independent.

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Where:

$k =$ specific number of successes
$n =$ number of trials
$p =$ probability of one success Mean (expected value) for the binomial distribution : The mean is given by:
$μ = E (X) = n \cdot p$

Variance equation for the binomial distribution: The variance is given by:
$σ^{2} = Var (X) = n \cdot p \cdot (1 - p)$
Standard deviation equation for the binomial distribution: The standard deviation is given by:
$σ = n \cdot p \cdot (1 - p)$

Conditions for Using the Binomial Distribution

Sidenote

Important to know!

To use the binomial distribution, all of the “BINS” conditions must be met.

“B” stands for binary outcomes. Each trial has exactly two outcomes: “success” or “failure.”
- Example: “Flipping heads is a success; flipping tails is a failure.”
- Example: “Rolling at least an $8$ on two dice is a success; rolling anything else is a failure.”
“I” stands for independent trials. The outcome of one trial does not affect the outcome of any other trial.
“N” stands for number of trials is fixed. You decide the number of trials in advance.
“S” stands for same probability of success. The probability of success $p$ is the same on every trial.

Practice problems (with calculator, but without software):

Example 1:

A baseball player’s probability of hitting a home run is $20%$ . What is the probability that the baseball player will hit a home run three times in a row?

Solution:

(spoiler)

The probability of one home run is $20% = 0.2$ . Because the trials are independent, multiply the probabilities:

$0.2 * 0.2 * 0.2 = 0. 2^{3} = 0.008$

Example 2:

What is the probability that the baseball player will not hit any home runs in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, the player bats $10$ times, so $n = 10$ . The probability of a home run is $p = 0.2$ . “No home runs” means $k = 0$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 0) = (0 10) 0. 2^{0} (1 - 0.2)^{10 - 0} = 1 * 1 * (0.8)^{10} = 0.1074$

So, the probability of no home runs in the next 10 at-bats is about $10.74%$ .

Example 3:

What is the probability that the baseball player will hit exactly one home run in his next ten times that he is up to bat?

Solution:

(spoiler)

Here, $n = 10$ , $p = 0.2$ , and “exactly one home run” means $k = 1$ . Recall the binomial probability formula:

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Substitute the values:

$P (X = 1) = (1 10) 0. 2^{1} (1 - 0.2)^{10 - 1} = 10 * 0.2 * (0.8)^{9} = 0.2684$

So, the probability of exactly one home run in the next 10 at-bats is about $26.84%$ .

Example 4:

What is the probability that the baseball player will hit more than one home run in his next 100 times up to bat?

Solution:

(spoiler)

To find $P (X > 1)$ directly, you’d need to add $P (X = 2) + P (X = 3) + \dots + P (X = 100)$ , which is not practical by hand. Instead, use the complement:

The opposite of “more than one” is “one or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ . From earlier parts, we have:
$P (X = 0) \approx 0.1074$
$P (X = 1) \approx 0.2684$ So,

$1 - (0.1074 + 0.2684) = 1 - 0.3758 = 0.6242$

Calculator Commands

binompdf $(n, p, x)$

Gives the probability of exactly $x$ successes in $n$ trials, where $p$ is the probability of success on one trial.

Example: binompdf $(3, 0.75, 2)$ gives the probability of exactly $2$ successes in $3$ trials when the probability of success on a single trial is $0.75$ .

binomcdf $(n, p, x)$

Gives the cumulative probability of $x$ or fewer successes in $n$ trials, where $p$ is the probability of success on a single trial.

Example: binomcdf $(3, 0.75, 2)$ gives the probability of $2$ or fewer successes (that is, $0$ , $1$ , or $2$ successes) in $3$ trials when the probability of success on a single trial is $0.75$ .

Practice problems (Part B and C with software):

Example 1:

A recent survey suggested that $85%$ of new cars purchased last year were gasoline cars and the rest were electric. If $100$ new cars that were purchased last year are chosen at random, calculate the mean and standard deviation and interpret them.

Solution:

(spoiler)

$E (X) = n * p = 100 * 0.85 = 85$

Interpretation: If you repeatedly took random samples of $100$ cars purchased last year, the average number of gasoline cars per sample would be about $85$ .

$σ = n * p * (1 - p) = 100 * 0.85 * (1 - 0.85) = 12.75 = 3.5707$

Interpretation: The count of gasoline cars in samples of size $100$ will typically differ from the mean of $85$ by about $3.5707$ cars.

Example 2:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that exactly $85$ of the cars were gasoline cars?

Solution:

(spoiler)

Use binompdf $(n, p, x)$ for “exactly $x$ successes.” Here:

$n = 100$
$p = 0.85$
$x = 85$ So we compute binompdf $(100, 0.85, 85)$ .

$binompdf (100, 0.85, 85) = 0.0968$

So, there is about a $9.68%$ chance of selecting exactly $85$ gasoline cars out of $100$ .

Example 3:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that no more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“No more than $85$ ” means $85$ or fewer, so use binomcdf $(n, p, x)$ with $x = 85$ . Here:

$n = 100$
$p = 0.85$
$x = 85$

$binomcdf (100, 0.85, 85) = 0.7151$

So, there is about a $71.51%$ chance that the sample contains $85$ or fewer gasoline cars.

Example 4:

If a random sample of $100$ new cars purchased last year was selected, what is the probability that more than $85$ of the cars were gasoline cars?

Solution:

(spoiler)

“More than $85$ ” is the complement of “ $85$ or fewer.” From the previous part:

$binomcdf (100, 0.85, 85) = 0.7151$

So,

$1 - binomcdf (100, 0.85, 85) = 1 - 0.7151 = 0.2849$

So, the probability of selecting more than $85$ gasoline cars is $0.2849$ .