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Magazine Chapter 24: Problem 156
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is (A) \(\frac{37}{256}\) (B) \(\frac{219}{256}\) (C) \(\frac{128}{256}\) (D) \(\frac{28}{256}\)
Short Answer
Expert verified
The probability of 2 successes is \( \frac{28}{256} \), which corresponds to option (D).
Step by step solution
01
Define the Binomial Distribution Parameters
A binomial distribution is defined by two parameters: the number of trials \( n \) and the probability of success \( p \). The mean \( \mu \) of a binomial distribution is given by \( \mu = np \) and the variance \( \sigma^2 \) is given by \( \sigma^2 = np(1-p) \). Here, we are given that \( \mu = 4 \) and \( \sigma^2 = 2 \).
02
Set Up the Equations for Mean and Variance
Using the formulas for mean and variance, we have two equations:1. \( np = 4 \)2. \( np(1-p) = 2 \).We can use these equations to find \( n \) and \( p \).
03
Solve for the Probability of Success \( p \)
From equation 1, express \( p \) in terms of \( n \):\[ p = \frac{4}{n} \].Substitute \( p \) into equation 2:\[ n \cdot \frac{4}{n} (1-\frac{4}{n}) = 2 \].Simplify to:\[ 4(1 - \frac{4}{n}) = 2 \]\[ 4 - \frac{16}{n} = 2 \]\[ \frac{16}{n} = 2 \]Solving for \( n \), we get \( n = 8 \).
04
Calculate the Probability \( p \)
Now that \( n = 8 \), substitute back to find \( p \):\[ p = \frac{4}{8} = \frac{1}{2} \].Thus, the probability of success \( p \) is \( \frac{1}{2} \).
05
Use the Binomial Formula for Probability
The probability of having exactly 2 successes (\( X = 2 \)) in a binomial distribution is given by the formula:\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]Substitute \( n = 8 \), \( r = 2 \), and \( p = \frac{1}{2} \):\[ P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{6} \]\[ = \binom{8}{2} \left(\frac{1}{2}\right)^8 \].
06
Calculate the Probability of 2 Successes
Compute \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \).Then, substitute into the probability formula:\[ P(X = 2) = 28 \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mean and Variance in Binomial Distribution
The world of binomial distribution offers fascinating insights into how probabilities are distributed over several trials. Two key components of this distribution are the mean and the variance. The mean of a binomial distribution is a measure of the average expected outcome, while the variance gives us an insight into how much the results can differ from this average. Both rely heavily on two parameters: the number of trials, typically denoted as \( n \), and the probability of success in each trial, denoted as \( p \).
These formulas tell us that if you know the mean and variance of a binomial distribution, you can work backward to find these underlying parameters. For example, given a mean of 4 and a variance of 2, these formulas allow us to derive the values of \( n \) and \( p \). By solving the formulas \( np = 4 \) and \( np(1-p) = 2 \), we unravel the specific characteristics of our distribution.
- Mean \( (\mu) \) is calculated using the formula: \( \mu = np \)
- Variance \( (\sigma^2) \) is calculated using the formula: \( \sigma^2 = np(1 - p) \)
These formulas tell us that if you know the mean and variance of a binomial distribution, you can work backward to find these underlying parameters. For example, given a mean of 4 and a variance of 2, these formulas allow us to derive the values of \( n \) and \( p \). By solving the formulas \( np = 4 \) and \( np(1-p) = 2 \), we unravel the specific characteristics of our distribution.
Probability of Success
In any binomial distribution, understanding the probability of success \( p \) is essential. It defines how likely it is that any single trial will succeed, thereby influencing our mean and variance significantly. For example, in the given exercise, solving for \( p \) starts with the equation \( np = 4 \). This equation relates the number of trials \( n \) to the probability of success \( p \).
Using the relationship \( p = \frac{4}{n} \), we can substitute \( p \) into the variance equation, \( np(1-p) = 2 \), to find \( n \). Once \( n \) is known, \( p \) is easily calculated. In this case, we found \( n = 8 \) and \( p = \frac{1}{2} \).
Having \( p = \frac{1}{2} \) indicates that each trial is equally likely to be a success or a failure, reflecting a balanced and classical binomial scenario.
Using the relationship \( p = \frac{4}{n} \), we can substitute \( p \) into the variance equation, \( np(1-p) = 2 \), to find \( n \). Once \( n \) is known, \( p \) is easily calculated. In this case, we found \( n = 8 \) and \( p = \frac{1}{2} \).
- The probability of success influences the shape of the distribution.
- It allows us to calculate other probabilities within the distribution.
Having \( p = \frac{1}{2} \) indicates that each trial is equally likely to be a success or a failure, reflecting a balanced and classical binomial scenario.
Binomial Probability Formula
The heart of calculating specific probabilities in a binomial distribution lies in the binomial probability formula. This formula helps in determining the likelihood of achieving a particular number of successes across several trials. The formula is given by:
\[P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\]
Where \( \binom{n}{r} \) is a binomial coefficient representing the number of combinations of \( n \) trials taken \( r \) at a time, and \( p \) is the success probability in each trial. For example, in our exercise, we found that the probability of exactly 2 successes was calculated by plugging the values into the formula:
\[P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^8\]
By solving \( \binom{8}{2} \) to 28, and calculating the power, we determined the probability as \( \frac{28}{256} \). This formula is a cornerstone in probability theory, unraveling the magic behind how likely certain outcomes are in terms of successes.
\[P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\]
Where \( \binom{n}{r} \) is a binomial coefficient representing the number of combinations of \( n \) trials taken \( r \) at a time, and \( p \) is the success probability in each trial. For example, in our exercise, we found that the probability of exactly 2 successes was calculated by plugging the values into the formula:
- \( n = 8 \)
- \( r = 2 \)
- \( p = \frac{1}{2} \)
\[P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^8\]
By solving \( \binom{8}{2} \) to 28, and calculating the power, we determined the probability as \( \frac{28}{256} \). This formula is a cornerstone in probability theory, unraveling the magic behind how likely certain outcomes are in terms of successes.
