Binomial Distribution: Videos & Practice Problems

In a binomial experiment, outcomes are limited to two possibilities, such as success or failure, across a fixed number of independent trials. The probability of success is denoted as p, while the probability of failure is q = 1 - p. The expected value (mean) is calculated using $\mu =n·p$, and the standard deviation is found using $\sigma =\surd (n·p·q)$. Understanding these concepts is crucial for analyzing probabilities in various scenarios.

The Binomial Experiment

Discrete random variables are essential in statistics, representing outcomes that cannot be subdivided further, such as the number of prizes in a raffle or the number of children in a household. A common scenario involving discrete random variables is a binomial experiment, which is characterized by having only two possible outcomes, often referred to as success and failure. For instance, in a coin flip, heads can be considered a success and tails a failure, though these terms do not inherently imply good or bad outcomes.

In a binomial experiment, the variable \( x \) denotes the number of successes, while the experiment consists of a series of independent trials. Independence means that the outcome of one trial does not influence the outcome of another. For example, flipping a coin multiple times results in independent outcomes. Each trial has a probability of success, denoted as \( p \), and a probability of failure, denoted as \( q \). The relationship between these probabilities is given by the equation \( q = 1 - p \). This means that if the probability of success is known, the probability of failure can be easily calculated.

To determine if an experiment qualifies as a binomial experiment, four criteria must be met: there must be two possible outcomes, a fixed number of trials, independent trials, and a consistent probability of success across trials. For example, if you flip a coin four times, you can count the number of times it lands on heads. Here, the outcomes are heads or tails, the number of trials is fixed at four, the trials are independent, and the probability of landing on heads (success) is 0.5, making the probability of tails (failure) also 0.5. Thus, this scenario meets all the criteria for a binomial experiment.

In contrast, consider an experiment where you draw marbles from a bag without replacement. If you pull out four marbles from a bag containing red and blue marbles, the outcomes are still red or blue, and the number of trials is fixed. However, since the outcome of one draw affects the probabilities of subsequent draws (because the total number of marbles decreases), the trials are not independent. Therefore, this scenario does not qualify as a binomial experiment.

Understanding these concepts is crucial for analyzing probabilities, means, and standard deviations in binomial experiments, which will be explored further in subsequent studies.

A basketball player normally has a 70% chance of making a free throw. The player shoots until finally making a basket, where $x$ is the number of shots they take. Is this a binomial experiment?

• A. Yes
• B. No, there are not 2 outcomes
• C. No, there are not a fixed # of trials
• D. No, the trials are not independent
• E. No, the probability of success is not the same for each trial

Answer: (C)

The Binomial Experiment Example 1

When analyzing whether a scenario qualifies as a binomial experiment, it is essential to evaluate specific criteria. In the case of rolling a standard six-sided die ten times, we begin by identifying the random variable, which is the outcome of each roll. The first step is to determine if there is a fixed number of trials. Since we are rolling the die ten times, this condition is satisfied.

Next, we assess the independence of the trials. The outcome of one roll does not influence the outcome of another, confirming that the trials are independent. The third criterion involves the probability of success. In this context, success can be defined in various ways, but there is an equal chance of landing on any of the six faces of the die.

However, the critical aspect to consider is whether there are only two possible outcomes. In this scenario, since the die can land on any of the six numbers (1, 2, 3, 4, 5, or 6), this condition fails. Therefore, rolling a die ten times does not constitute a binomial experiment.

In a modified scenario where we define the random variable as the number of times a '2' is rolled, we can re-evaluate the criteria. We still have a fixed number of trials (ten rolls), the trials remain independent, and there is an equal probability of success (rolling a '2'). Now, we must consider the outcomes: we can either roll a '2' (success) or roll one of the other five numbers (failure). This definition provides exactly two possible outcomes, thus satisfying all the conditions for a binomial experiment.

In summary, while the first example of rolling a die ten times does not meet the criteria for a binomial experiment due to the multiple outcomes, redefining the random variable to count the number of times a specific outcome occurs (like rolling a '2') transforms it into a valid binomial experiment.

The Binomial Experiment Example 2

In this scenario, we analyze a pharmaceutical company's claim regarding a new vaccine's effectiveness, specifically its ability to confer immunity. The company states that the probability of gaining immunity from the vaccine is 80%. When administered to 100 patients, 83 of them achieved immunity. To determine if this situation represents a binomial experiment, we must verify four key conditions.

First, we confirm that there are two possible outcomes: a patient either gains immunity or does not. This satisfies the requirement for binary outcomes. Second, the experiment involves a fixed number of trials, as the vaccine is given to exactly 100 patients. Third, the trials are independent; the outcome for one patient does not influence the outcome for another. Lastly, the probability of success remains constant at 80% for each trial, fulfilling the condition of equal probability.

Since all conditions are met, we can classify this as a binomial experiment. Next, we identify the relevant parameters:

The variable n represents the number of trials, which in this case is 100, as the vaccine was administered to 100 patients. The probability of success, denoted as p, is given as 80%, which can be expressed as a decimal: p = 0.8. The probability of failure, q, is calculated using the relationship q = 1 - p. Therefore, q = 1 - 0.8 = 0.2.

Finally, the variable x represents the number of successes, which is the number of patients who gained immunity. In this case, x = 83.

In summary, for this binomial experiment, we have:

• n = 100 (number of trials)
• p = 0.8 (probability of success)
• q = 0.2 (probability of failure)
• x = 83 (number of successes)

Calculating Probabilities in a Binomial Distribution

In a binomial experiment, you often need to calculate the probability of achieving a specific number of successes in a series of independent trials. For instance, if you flip a coin 10 times, you might want to find the probability of getting exactly 4 heads. This can be accomplished using a straightforward formula rather than complex tables or exhaustive outcome lists.

To illustrate this, consider a scenario where you have a bag containing red and blue marbles. You draw one marble, replace it, and repeat this process four times. If the probability of drawing a red marble is 80%, you can determine the probability of drawing exactly 3 red marbles.

First, identify the key components of the experiment: the number of trials (n), the probability of success (p), the probability of failure (q), and the desired number of successes (x). In this case, n is 4 (the number of draws), p is 0.8 (the probability of drawing a red marble), and q is 0.2 (the probability of drawing a blue marble, calculated as 1 - p). The desired number of successes, x, is 3 (the number of red marbles you want to draw).

Next, to find the probability of drawing exactly 3 red marbles, you need to consider the different ways this can occur. The probability of drawing 3 red marbles and 1 blue marble can be expressed as:

\[ P(X = 3) = \binom{n}{x} p^x q^{n-x} \]

Here, \[ \binom{n}{x} \] represents the number of combinations of n trials taken x at a time, which can be calculated using the formula:

\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \]

For our example, this becomes:

\[ \binom{4}{3} = \frac{4!}{3!1!} = 4 \]

Now, substituting the values into the probability formula gives:

\[ P(X = 3) = 4 \cdot (0.8)^3 \cdot (0.2)^1 \]

Calculating this yields:

\[ P(X = 3) = 4 \cdot 0.512 \cdot 0.2 = 0.4096 \]

Thus, the probability of drawing exactly 3 red marbles is approximately 0.41, or 41%.

In summary, the general formula for calculating the probability of obtaining x successes in n trials is:

\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]

By identifying the variables and applying this formula, you can efficiently calculate probabilities in binomial experiments.

You take a 6-question quiz with True/False questions. What is the probability of getting all 6 questions correct by simply guessing?

• A. 0.015%
• B. 0.50%
• C. 5.6%
• D. 1.56%

Answer: (D)

National surveys indicate that 36% of people have been in a car accident in the last 5 years. If you randomly sample 10 people, how likely is that exactly 4 have had an accident in the last 5 years?

• A. 0.0767
• B. 0.242
• C. 0.00527
• D. 0.00129

Answer: (B)

Mean & Standard Deviation of Binomial Distribution

In the study of binomial distributions, two key quantities of interest are the mean (or expected value) and the standard deviation. These can be calculated using simplified formulas that rely solely on the number of trials and the probability of success, rather than more complex summation equations.

For a binomial distribution, the expected value (denoted as μ) can be calculated using the formula:

\[\mu = n \times p\]

where:

• n = number of trials (e.g., the number of customers surveyed)
• p = probability of success (e.g., the likelihood that a customer will purchase the product)

For example, if a company expects that 60% of customers will purchase a product after a demonstration, and 10 customers are randomly selected, the expected number of purchases can be calculated as:

\[\mu = 10 \times 0.6 = 6\]

This means that, on average, 6 out of the 10 customers are expected to buy the product.

To find the standard deviation (denoted as σ) of a binomial distribution, the formula is:

\[\sigma = \sqrt{n \times p \times q}\]

where:

• q = probability of failure (calculated as q = 1 - p)

Continuing with the previous example, if the probability of success is 0.6, then the probability of failure is:

\[q = 1 - 0.6 = 0.4\]

Now, substituting the values into the standard deviation formula gives:

\[\sigma = \sqrt{10 \times 0.6 \times 0.4} = \sqrt{2.4} \approx 1.55\]

This indicates that the standard deviation of the number of customers expected to purchase the product is approximately 1.55 customers.

Additionally, the variance (denoted as σ²) can be derived from the standard deviation, as it is simply the square of the standard deviation:

\[\sigma^2 = n \times p \times q = 2.4\]

In summary, for a binomial distribution, the mean and standard deviation can be efficiently calculated using these straightforward formulas, allowing for quick analysis of expected outcomes and variability in scenarios involving discrete random variables.

Based on historical weather data in a certain city, about 62% of the days are cloudy. Find the mean, standard deviation, and variance for the number of cloudy days in a 30-day month.

• A. Mean = 18.6 days; Standard deviation = 2.66 days; Variance = $7.08days^2$
• B. Mean = 18.6 days; Standard deviation = 3.38 days; Variance = $11.4days^2$
• C. Mean = 11.4 days; Standard deviation = 2.66 days; Variance = $7.08days^2$
• D. Mean = 11.4 days; Standard deviation = 3.38 days; Variance = $11.4days^2$

Answer: (A)

Finding Multiple Probabilities in Binomial Distribution

In probability theory, particularly when dealing with binomial distributions, it is common to calculate the likelihood of a certain number of successes in a given number of trials. When tasked with finding the probability of a range of successes, such as between two numbers or at least a certain number, the approach involves summing the probabilities of individual outcomes.

For instance, consider a scenario where a marketing company sends promotional emails to 10 customers, with a 40% probability (p) that any given customer will open the email. The probability of failure (q) is therefore 60% (1 - p). To find the probability that between 0 and 2 customers open the email, you would calculate the probabilities for exactly 0, 1, and 2 customers opening the email and then sum these probabilities. This is expressed mathematically as:

\[P(X = k) = \binom{n}{k} p^k q^{n-k}\]

where \( n \) is the number of trials, \( k \) is the number of successes, \( p \) is the probability of success, and \( q \) is the probability of failure. For our example:

1. For \( k = 0 \):\[P(X = 0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = 0.006\]2. For \( k = 1 \):\[P(X = 1) = \binom{10}{1} (0.4)^1 (0.6)^{9} = 0.040\]3. For \( k = 2 \):\[P(X = 2) = \binom{10}{2} (0.4)^2 (0.6)^{8} = 0.121\]

Adding these probabilities gives:

\[P(0 \leq X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.006 + 0.040 + 0.121 = 0.167\]

This indicates a 16.7% chance that between 0 and 2 customers will open the email.

In another scenario, if you need to find the probability that at least 3 customers open the email, you can use the complement rule. Instead of calculating the probabilities for 3 through 10 directly, you can find the probability of the opposite event (less than or equal to 2) and subtract it from 1:

\[P(X \geq 3) = 1 - P(X \leq 2)\]

Since we already calculated \( P(X \leq 2) = 0.167 \), we find:

\[P(X \geq 3) = 1 - 0.167 = 0.833\]

This means there is an 83.3% chance that at least 3 customers will open the email. Utilizing the complement can significantly simplify calculations, especially when dealing with larger ranges of outcomes.

A gardener plants 8 seeds, and each has a 65% probability of germinating successfully. Find the probability that less than 4 seeds germinate successfully.

• A. 0.0266
• B. 0.0252
• C. 0.1062
• D. 0.0084

Answer: (C)

Finding Binomial Probabilities Using TI-84

In a binomial experiment, we often need to calculate the probability of a certain number of successes, which can be complex, especially with a large number of trials. A practical approach to simplify these calculations is to use a graphing calculator, such as the TI-84, which can efficiently compute binomial probabilities.

Consider a manufacturing company that produces light bulbs with a defect rate of 5%. If the company produces 1,000 light bulbs in a day, we can model the number of defective light bulbs as a binomial distribution where the number of trials (n) is 1,000 and the probability of success (p) is 0.05. In this context, a "success" refers to producing a defective light bulb.

To find the probability of exactly 50 defective light bulbs, we use the binomial probability formula, which can be accessed through the binomPDF function on the TI-84. This function requires three inputs: the number of trials (n = 1000), the probability of success (p = 0.05), and the number of successes (x = 50). After entering these values, the calculator provides the probability, which in this case is approximately 0.058, indicating a 5.8% chance of exactly 50 defective light bulbs.

For cumulative probabilities, such as finding the probability of having at most 50 defective light bulbs, we utilize the binomCDF function. This function calculates the probability of having a number of successes less than or equal to a specified value. By entering the same values (n = 1000, p = 0.05, x = 50), the calculator yields a probability of 0.538, or a 53.8% chance of having at most 50 defective light bulbs.

If we want to determine the probability of having less than 50 defective light bulbs, we can still use the binomCDF function, but we need to adjust our x value to 49. This adjustment allows us to find the probability of having 49 or fewer defective light bulbs. The resulting probability is approximately 0.48, indicating a 48% chance of having fewer than 50 defective light bulbs. This result can also be verified by subtracting the probability of exactly 50 defective light bulbs from the cumulative probability of at most 50 defective light bulbs.

Utilizing the TI-84 for binomial probability calculations streamlines the process, making it easier to handle complex scenarios without extensive manual computation. Practice with these functions will enhance your proficiency in statistical analysis.

A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(A) What is the probability that exactly 450 nesting attempts are successful?

• A. $3.54\times {10}^{-78}$
• B. $0.70$
• C. $3.3\times {10}^{-27}$
• D. $1$

Answer: (C)

A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(B) What is the probability that less than 330 attempts are successful?

• A. $0.02$
• B. $0.005$
• C. $0.98$
• D. $0.03$

Answer: (A)

A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(C) What is the probability that at least 330 nesting attempts are successful?

• A. $0.02$
• B. $0.005$
• C. $0.98$
• D. $0.03$

Answer: (C)

A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(D) What is the probability that 330 – 370 nesting attempts are successful?

• A. $0.98$
• B. $0.96$
• C. $0.02$
• D. 0.04

Answer: (B)

Finding Binomial Probabilities Using TI-84 Example 1

In this scenario, a medical researcher is evaluating the effectiveness of a new vaccine, which has been established to be 90% effective in preventing infection when administered correctly. The study involves a binomial experiment, characterized by two possible outcomes: success (the vaccine is effective) or failure (the vaccine is ineffective). With a fixed number of trials—10 patients in this case—and a consistent probability of success (p = 0.9), the independence of each patient's outcome can be assumed due to the random sampling method.

To determine the probability that the vaccine was effective for seven or fewer patients, we need to calculate \( P(X \leq 7) \). Instead of calculating the probabilities for each individual outcome from 0 to 7, which would be cumbersome, we can utilize the binomial cumulative distribution function (CDF). This function allows us to find the cumulative probability directly.

Using a calculator, specifically the TI-84, we access the distribution menu and select the binomial CDF option. We input the number of trials (10), the probability of success (0.9), and the value of interest (7). After processing this, we find that the probability is approximately 0.07, or 7%. This indicates that there is a 7% chance that the vaccine is effective for seven or fewer patients.

Next, we assess whether this result is statistically significant. Statistical significance is determined by comparing the calculated probability against predefined significance levels, typically set at 1%, 5%, and 10%. In this case, since our probability of 7% exceeds both the 1% and 5% significance levels, we conclude that the result is not statistically significant at these thresholds. However, when comparing to the 10% significance level, since 7% is less than 10%, we find that the result is statistically significant. This suggests that further investigation into the vaccine's effectiveness may be warranted.

Finding Binomial Probabilities Using TI-84 Example 2

In this scenario, we are analyzing a binomial experiment involving customer service calls, where the probability of a satisfied customer is 90% (or 0.9). The total number of calls made is 20, which establishes our number of trials, denoted as \( n = 20 \). The outcomes of each call are independent, leading us to use the binomial distribution to calculate probabilities.

To find the probability of 15 or fewer satisfied customers, we denote this as \( P(X \leq 15) \). Instead of calculating the probabilities for each individual outcome from 0 to 15, we can utilize the binomial cumulative distribution function (binom CDF) on a calculator. By inputting the values \( n = 20 \), \( p = 0.9 \), and \( x = 15 \), we find that the probability is approximately 0.04, or 4%. This indicates that there is a 4% chance of having 15 or fewer satisfied customers out of 20 calls.

Next, we apply the range rule of thumb to assess whether observing 15 or fewer satisfied customers is statistically unusual. According to this rule, an outcome is considered significant if it lies two or more standard deviations away from the mean. The mean of a binomial distribution is calculated as \( \mu = n \times p \). Here, \( \mu = 20 \times 0.9 = 18 \).

The standard deviation is determined using the formula \( \sigma = \sqrt{n \times p \times q} \), where \( q = 1 - p \). Thus, \( q = 0.1 \). Plugging in the values, we calculate:

\( \sigma = \sqrt{20 \times 0.9 \times 0.1} \approx 1.3 \).

To evaluate if 15 is significantly low, we calculate two standard deviations below the mean:

\( 2 \times \sigma \approx 2 \times 1.3 = 2.6 \).

Subtracting this from the mean gives:

\( 18 - 2.6 = 15.4 \).

Since 15 is less than 15.4, it indicates that 15 is more than two standard deviations away from the mean. Therefore, the manager should indeed be concerned about a potential decline in service quality, as observing 15 or fewer satisfied customers is statistically unusual.

Here’s what students ask on this topic:

What are the criteria for a binomial experiment?

A binomial experiment must satisfy four key criteria: (1) There are a fixed number of trials, denoted as n. (2) Each trial has only two possible outcomes: success or failure. (3) The probability of success, p, remains constant for each trial. (4) The trials are independent, meaning the outcome of one trial does not affect the others. For example, flipping a coin 10 times to count the number of heads is a binomial experiment because it meets all these criteria. Understanding these conditions is essential to correctly identify and analyze binomial distributions in probability and statistics.

How do you calculate the probability of exactly x successes in a binomial experiment?

The probability of exactly x successes in a binomial experiment is calculated using the binomial probability formula:

$P(X=x)=\frac{n!}{x!(n-x)!}\cdot p^x\cdot q^{n-x}$

Here, n is the number of trials, p is the probability of success, q is the probability of failure (q = 1 - p), and x is the desired number of successes. For example, if you flip a coin 5 times (n = 5) and want the probability of getting exactly 3 heads (x = 3, p = 0.5), you would substitute these values into the formula to calculate the probability.

How do you find the mean and standard deviation of a binomial distribution?

The mean (μ) and standard deviation (σ) of a binomial distribution can be calculated using the following formulas:

Mean: $\mu =n\cdot p$

Standard Deviation: $\sigma =\sqrt{n\cdot p\cdot q}$

Here, n is the number of trials, p is the probability of success, and q is the probability of failure (q = 1 - p). For example, if a coin is flipped 10 times (n = 10) with a probability of heads being 0.5 (p = 0.5), the mean would be $\mu =10\cdot 0.5=5$, and the standard deviation would be $\sigma =\sqrt{10\cdot 0.5\cdot 0.5}=1.58$.

What is the difference between binomial PDF and binomial CDF?

Binomial PDF (Probability Density Function) calculates the probability of exactly x successes in a binomial experiment. For example, it answers questions like, "What is the probability of getting exactly 3 heads in 10 coin flips?"

Binomial CDF (Cumulative Distribution Function), on the other hand, calculates the probability of getting up to a certain number of successes (e.g., x ≤ a specific value). It answers questions like, "What is the probability of getting 3 or fewer heads in 10 coin flips?"

In practice, PDF is used for exact probabilities, while CDF is used for cumulative probabilities. Both functions are available on calculators like the TI-84, making it easier to compute these probabilities for large datasets.

How do you use a TI-84 calculator to find binomial probabilities?

To find binomial probabilities on a TI-84 calculator:

1. For exact probabilities (Binomial PDF): Press 2nd → VARS to access the distribution menu. Scroll to binompdf, enter the number of trials (n), probability of success (p), and desired successes (x), then press ENTER.

2. For cumulative probabilities (Binomial CDF): Follow the same steps but select binomcdf. Enter n, p, and the maximum number of successes for the range, then press ENTER.

For example, to find the probability of at most 3 successes in 10 trials with p = 0.5, use binomcdf(10, 0.5, 3). The calculator will return the cumulative probability.