Binomial Distribution: Videos & Practice Problems

Discrete random variables are essential in probability and 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.

In a binomial experiment, there are four key criteria to consider:

1. Two Possible Outcomes: Each trial results in one of two outcomes, such as heads or tails.
2. Fixed Number of Trials: The experiment consists of a predetermined number of trials, denoted as n. For example, flipping a coin four times means n = 4.
3. Independent Trials: The outcome of one trial does not influence the outcome of another. Each coin flip is independent of the others.
4. Constant Probability of Success: Each trial has a consistent probability of success, represented as p, and the probability of failure, q, is calculated as q = 1 - p.

For example, if you flip a coin four times and count the number of heads, you can determine the values of n, p, q, and x (the number of successes). In this case, n = 4, p = 0.5 (for heads), q = 0.5 (for tails), and if you get heads three times, then x = 3.

However, not all experiments meet the criteria for a binomial experiment. For instance, if you draw marbles from a bag without replacement, the trials are not independent because removing a marble affects the probabilities of subsequent draws. In this case, even though there are two outcomes (red or blue), the lack of independence disqualifies it as a binomial experiment.

Understanding these principles is crucial for analyzing binomial experiments and calculating probabilities, means, and standard deviations in future studies.

When analyzing whether a scenario qualifies as a binomial experiment, it is essential to evaluate specific criteria. In the context of rolling a six-sided die, we can explore two different setups to illustrate these principles.

In the first scenario, you roll a six-sided die 10 times and record the outcome of each roll. Here, we assess the conditions for a binomial experiment:

• Fixed Number of Trials: You are rolling the die 10 times, which satisfies this condition.
• Independence of Trials: The outcome of one roll does not affect the others, confirming independence.
• Equal Chance of Success: Each face of the die has an equal probability of landing face up.
• Two Possible Outcomes: This is where the criteria fails. Since there are six possible outcomes (1, 2, 3, 4, 5, or 6), this scenario does not meet the requirement of having only two outcomes. Therefore, this is not a binomial experiment.

In the second scenario, you again roll the die 10 times, but this time you define the random variable as the number of times you roll exactly a two. Evaluating the same criteria:

• Fixed Number of Trials: You still roll the die 10 times.
• Independence of Trials: Each roll remains independent of the others.
• Equal Chance of Success: The probability of rolling a two is consistent across trials.
• Two Possible Outcomes: Now, the outcomes can be categorized as either rolling a two (success) or rolling one of the other five numbers (failure). This meets the requirement for two possible outcomes.

Since all conditions are satisfied in this second scenario, it qualifies as a binomial experiment. Understanding these criteria is crucial for correctly identifying binomial experiments in various contexts.

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 individuals 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 binary outcome is essential for a binomial experiment. Second, the number of trials is fixed, 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 requirement for equal probability.

Since all conditions are satisfied, we conclude that this is indeed a binomial experiment. We can now identify the relevant parameters:

Let \( n \) represent 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, we define \( x \), the discrete random variable representing the number of successes. Here, \( x \) is the number of patients who gained immunity, which is 83. Thus, we summarize the parameters of this binomial experiment as follows:

- \( n = 100 \) (number of trials)

- \( p = 0.8 \) (probability of success)

- \( q = 0.2 \) (probability of failure)

- \( x = 83 \) (number of successes)

This analysis provides a clear understanding of the binomial experiment's structure and the associated probabilities, allowing for further statistical evaluation if needed.

In probability theory, a binomial experiment consists of a fixed number of independent trials, each with two possible outcomes: success and failure. To calculate the probability of achieving an exact number of successes, we can utilize a straightforward formula that incorporates key variables from the experiment.

Consider a scenario where you draw marbles from a bag containing red and blue marbles. If you draw one marble, replace it, and repeat this process four times, you are conducting a binomial experiment. In this case, the probability of drawing a red marble (success) is 80% (or 0.8), while the probability of drawing a blue marble (failure) is 20% (or 0.2).

To solve for the probability of drawing exactly three red marbles (successes), we first identify the essential variables:

• n: the total number of trials (4 in this case).
• p: the probability of success (0.8 for red marbles).
• q: the probability of failure (0.2 for blue marbles, calculated as 1 - p).
• x: the desired number of successes (3 red marbles).

To find the probability of exactly three successes, we apply the binomial probability formula:

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

Here, \(\binom{n}{x}\) represents the number of combinations of n trials taken x at a time, calculated as:

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

For our example, this becomes:

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

Next, we substitute the values into the formula:

\[P(X = 3) = \binom{4}{3} (0.8)^3 (0.2)^{4-3} = 4 \cdot (0.8)^3 \cdot (0.2)^1\]

Calculating this gives:

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

Thus, the probability of drawing exactly three red marbles is approximately 0.41, or 41%. This method can be generalized for any binomial experiment by identifying the relevant variables and applying the binomial probability formula.

In the study of binomial distributions, two key statistical measures are the mean (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 complex summations.

For a binomial distribution, the mean (μ) can be determined using the formula:

$$\mu = n \times p$$

where n represents the number of trials and p is the probability of success. 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 indicates that, on average, 6 out of the 10 customers are expected to buy the product.

To find the standard deviation (σ) of the binomial distribution, the formula used is:

$$\sigma = \sqrt{n \times p \times q}$$

Here, q is the probability of failure, which can be calculated as q = 1 - p. In this case, since the probability of success is 0.6, the probability of failure is:

$$q = 1 - 0.6 = 0.4$$

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 means the standard deviation of the number of customers expected to purchase the product is approximately 1.55.

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

$$\sigma^2 = n \times p \times q = 2.4$$

Variance is often expressed without units, but it represents the spread of the distribution. In summary, for the given scenario, the mean is 6 customers, the standard deviation is approximately 1.55 customers, and the variance is 2.4 customers squared.

In this scenario, a bank is analyzing the monthly usage of its online platform and discovers that 20% of users are inactive, meaning they do not log into their accounts. The bank plans to randomly sample 50 customer accounts to determine the expected number of active users who log in at least once a month. This situation can be modeled as a binomial experiment, where each account can either be active or inactive.

The expected value, often referred to as the mean (denoted by the symbol μ), can be calculated using the formula:

μ = n * p

where n is the number of trials (in this case, 50 customer accounts) and p is the probability of success (the proportion of active users). Since 20% of users are inactive, the probability of an account being active is:

p = 1 - q = 1 - 0.2 = 0.8

Thus, the expected number of active users is:

μ = 50 * 0.8 = 40

This means that, on average, the bank can expect to find 40 active users in a sample of 50 accounts.

To assess the variability of the number of active users, the standard deviation (denoted by σ) can be calculated using the formula:

σ = √(n * p * q)

Substituting the known values:

σ = √(50 * 0.8 * 0.2) = √(8) ≈ 2.83

This standard deviation indicates that the number of active users in repeated samples of 50 accounts will typically vary by about 2.83 users from the expected value of 40. Understanding these concepts of expected value and standard deviation is crucial for analyzing user engagement and making informed decisions based on statistical data.

In probability theory, particularly when dealing with binomial distributions, it is common to calculate the likelihood of a range of successes rather than just an exact number. This involves using the binomial probability formula, which is expressed as:

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

where:

• $$ n $$ is the number of trials,
• $$ k $$ is the number of successes,
• $$ p $$ is the probability of success on a single trial, and
• $$ \binom{n}{k} $$ is the binomial coefficient, representing the number of ways to choose $$ k $$ successes from $$ n $$ trials.

When tasked with finding the probability of a range of successes, such as "between zero and two customers opening an email," you would sum the probabilities of each individual outcome within that range. For example, this would involve calculating:

$$ P(X = 0) + P(X = 1) + P(X = 2) $$

In a practical scenario, if a marketing company sends an email to 10 customers (where $$ n = 10 $$) and the probability of a customer opening the email is 40% (thus, $$ p = 0.4 $$ and $$ q = 0.6 $$), the calculations for the probabilities would be as follows:

1. For zero customers opening the email:

$$ P(X = 0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = 0.006 $$

2. For one customer opening the email:

$$ P(X = 1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 0.040 $$

3. For two customers opening the email:

$$ P(X = 2) = \binom{10}{2} (0.4)^2 (0.6)^8 = 0.121 $$

Adding these probabilities together gives:

$$ P(0 \leq X \leq 2) = 0.006 + 0.040 + 0.121 = 0.167 $$

This indicates a 16.7% chance that between zero and two customers will open the email.

For scenarios where you need to find the probability of "at least" a certain number of successes, such as "at least three customers opening the email," it can be more efficient to use the complement rule. This means calculating the probability of the opposite event (in this case, "less than or equal to two customers opening the email") and subtracting it from one:

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

Since we already calculated $$ P(X \leq 2) $$ as 0.167, we can find:

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

This results in an 83.3% chance that at least three customers will open the email. Understanding how to manipulate these probabilities and apply the complement rule can significantly simplify calculations in binomial probability scenarios.

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 producing light bulbs with a defect rate of 5%. If the company produces 1,000 light bulbs in a day, we can define each light bulb as a trial in our binomial experiment. Here, the probability of success (defective light bulb) is denoted as \( p = 0.05 \).

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 \( x = 50 \). The command structure is:

binomPDF(n, p, x)

After entering these values, the calculator provides the probability, which in this case is approximately 0.058, indicating a 5.8% chance that exactly 50 light bulbs are defective.

Next, to find the probability of having at most 50 defective light bulbs (i.e., \( x \leq 50 \)), we utilize the binomCDF function. This function calculates cumulative probabilities, allowing us to find the probability of having 50 or fewer successes. The command structure remains similar:

binomCDF(n, p, x)

By entering \( n = 1000 \), \( p = 0.05 \), and \( x = 50 \), the calculator yields a probability of approximately 0.538, or a 53.8% chance of having at most 50 defective light bulbs.

For the probability of having less than 50 defective light bulbs (i.e., \( x < 50 \)), we can still use the binomCDF function, but we need to adjust our \( x \) value to 49. This means we are looking for \( P(X \leq 49) \). The command would be:

binomCDF(n, p, 49)

Upon calculation, this results in a probability of approximately 0.48, indicating a 48% chance that fewer than 50 light bulbs are defective. This 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:

P(X < 50) = P(X \leq 50) - P(X = 50)

Thus, using the TI-84 calculator streamlines the process of calculating binomial probabilities, making it easier to handle complex scenarios in binomial experiments.

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 trial is assumed due to the random sampling of patients.

To determine the probability that the vaccine was effective for seven or fewer patients, we denote this as P(X ≤ 7). Instead of calculating the probabilities for each individual outcome (from 0 to 7), which would be cumbersome, the researcher utilizes the binomial cumulative distribution function (CDF) on a TI-84 calculator. By accessing the distribution menu and selecting the binomcdf function, the researcher inputs the number of trials (10), the probability of success (0.9), and the desired x value (7). The resulting probability is approximately 0.07, or 7%, indicating a 7% chance that the vaccine was effective for seven or fewer patients.

In the second part of the analysis, the researcher assesses whether this result is statistically significant. Statistical significance implies that the observed outcome is unlikely enough to warrant further investigation. The researcher compares the calculated probability (7%) against various significance levels: 1%, 5%, and 10%. Since 7% exceeds both the 1% and 5% significance levels, these results are not considered statistically significant. However, when compared to the 10% significance level, 7% is less than 10%, suggesting that the result is statistically significant and merits further examination.

In this scenario, a company is analyzing the quality of its customer service calls, where historical data indicates that 90% of calls result in satisfied customers. This situation exemplifies a binomial experiment, characterized by a fixed number of trials (20 calls), two possible outcomes (satisfied or unsatisfied), a constant probability of success (0.9), and independent trials.

To determine the probability of 15 or fewer satisfied customers, we denote the number of satisfied customers as \(X\). We seek \(P(X \leq 15)\). Instead of calculating individual probabilities for \(X = 15, 14, 13, \ldots\), we can utilize the binomial cumulative distribution function (binom CDF) on a calculator. For this problem, we input the number of trials \(n = 20\), the probability of success \(p = 0.9\), and the value \(x = 15\). The resulting probability is approximately 0.04, or 4%, indicating a low likelihood of having 15 or fewer satisfied customers.

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. We first calculate the mean (\(\mu\)) of the distribution using the formula:

\[\mu = n \times p = 20 \times 0.9 = 18\]

Next, we compute the standard deviation (\(\sigma\)) using the formula:

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

Here, \(q\) represents the probability of failure, calculated as \(q = 1 - p = 0.1\). Thus, we find:

\[\sigma = \sqrt{20 \times 0.9 \times 0.1} \approx 1.3\]

To determine if 15 is significantly unusual, we calculate two standard deviations:

\[2 \times \sigma \approx 2 \times 1.3 = 2.6\]

Subtracting this from the mean gives:

\[\mu - 2\sigma = 18 - 2.6 = 15.4\]

Since 15 is less than 15.4, it indicates that 15 is more than two standard deviations below 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.