In a binomial distribution, the mean is \(\dfrac{2}{3}\) and variance is \(\dfrac{5}{9}\) . What is the probability that random variable D = 2?
Calculating Probability in Binomial Distribution
This question asks for the probability of a specific outcome (random variable D = 2) in a binomial distribution, given its mean and variance. To solve this, we first need to determine the parameters of the binomial distribution: the number of trials (\(n\)) and the probability of success in a single trial (\(p\)). The probability of failure (\(q\)) is then \(1-p\).
Understanding Binomial Distribution Parameters
A binomial distribution is defined by two parameters: \(n\) (number of independent Bernoulli trials) and \(p\) (probability of success on each trial). The probability of getting exactly \(k\) successes in \(n\) trials is given by the formula:
\[P(X=k) = \binom{n}{k} p^k q^{n-k}\]
where \(\binom{n}{k} = \dfrac{n!}{k!(n-k)!}\) is the binomial coefficient, and \(q = 1-p\).
The mean (\(\mu\)) and variance (\(\sigma^2\)) of a binomial distribution are related to \(n\) and \(p\) by the following formulas:
- Mean: \(\mu = np\)
- Variance: \(\sigma^2 = npq\)
Finding the Binomial Parameters (n, p, q)
We are given the mean and variance:
- Mean (\(np\)) = \(\dfrac{2}{3}\)
- Variance (\(npq\)) = \(\dfrac{5}{9}\)
We can use these two equations to find \(p\), \(q\), and \(n\).
Divide the variance by the mean:
\[\dfrac{npq}{np} = \dfrac{5/9}{2/3}\]
The \(np\) terms cancel out on the left side, leaving \(q\):
\[q = \dfrac{5}{9} \times \dfrac{3}{2} = \dfrac{15}{18} = \dfrac{5}{6}\]
Now that we have \(q\), we can find \(p\) using \(p = 1-q\):
\[p = 1 - \dfrac{5}{6} = \dfrac{6-5}{6} = \dfrac{1}{6}\]
Finally, we can find \(n\) using the mean formula \(np = \dfrac{2}{3}\) and the value of \(p = \dfrac{1}{6}\):
\[n \times \dfrac{1}{6} = \dfrac{2}{3}\]
Multiply both sides by 6:
\[n = \dfrac{2}{3} \times 6 = \dfrac{12}{3} = 4\]
So, the parameters of the binomial distribution are \(n=4\), \(p=\dfrac{1}{6}\), and \(q=\dfrac{5}{6}\).
Calculating the Probability \(P(D=2)\)
We need to find the probability that the random variable D equals 2. This means we need to calculate \(P(X=2)\) using the binomial probability formula with \(n=4\), \(k=2\), \(p=\dfrac{1}{6}\), and \(q=\dfrac{5}{6}\).
\[P(D=2) = \binom{4}{2} \left(\dfrac{1}{6}\right)^2 \left(\dfrac{5}{6}\right)^{4-2}\]
\[P(D=2) = \binom{4}{2} \left(\dfrac{1}{6}\right)^2 \left(\dfrac{5}{6}\right)^2\]
First, calculate the binomial coefficient \(\binom{4}{2}\):
\[\binom{4}{2} = \dfrac{4!}{2!(4-2)!} = \dfrac{4!}{2!2!} = \dfrac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \dfrac{24}{4} = 6\]
Now substitute this back into the probability formula:
\[P(D=2) = 6 \times \left(\left(\dfrac{1}{6}\right)^2\right) \times \left(\left(\dfrac{5}{6}\right)^2\right)\]
\[P(D=2) = 6 \times \left(\dfrac{1}{36}\right) \times \left(\dfrac{25}{36}\right)\]
Multiply the terms:
\[P(D=2) = \dfrac{6 \times 1 \times 25}{36 \times 36}\]
\[P(D=2) = \dfrac{150}{1296}\]
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 6:
\[\dfrac{150 \div 6}{1296 \div 6} = \dfrac{25}{216}\]
So, the probability that random variable D = 2 is \(\dfrac{25}{216}\).
Summary of Calculations
| Parameter | Calculation | Value |
|---|---|---|
| Mean (\(np\)) | Given | \(\dfrac{2}{3}\) |
| Variance (\(npq\)) | Given | \(\dfrac{5}{9}\) |
| \(q\) | \(\dfrac{\text{Variance}}{\text{Mean}} = \dfrac{5/9}{2/3}\) | \(\dfrac{5}{6}\) |
| \(p\) | \(1-q = 1 - \dfrac{5}{6}\) | \(\dfrac{1}{6}\) |
| \(n\) | \(\dfrac{\text{Mean}}{p} = \dfrac{2/3}{1/6}\) | \(4\) |
| \(P(D=2)\) | \(\binom{4}{2} \left(\dfrac{1}{6}\right)^2 \left(\dfrac{5}{6}\right)^2\) | \(\dfrac{25}{216}\) |
Revision Table: Binomial Distribution Concepts
| Concept | Definition/Formula |
|---|---|
| Binomial Distribution | A discrete probability distribution for the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. |
| Parameters (\(n, p\)) | \(n\): Number of trials \(p\): Probability of success on a single trial \(q\): Probability of failure on a single trial (\(q=1-p\)) |
| Mean (\(\mu\)) | Expected number of successes: \(\mu = np\) |
| Variance (\(\sigma^2\)) | Measure of spread: \(\sigma^2 = npq\) |
| Probability Mass Function (PMF) | \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) for \(k=0, 1, \ldots, n\) |
| Binomial Coefficient (\(\binom{n}{k}\)) | The number of ways to choose \(k\) successes from \(n\) trials: \(\dfrac{n!}{k!(n-k)!}\) |
Additional Information: Properties of Binomial Distribution
The binomial distribution is a fundamental concept in probability and statistics. It applies to situations where an experiment is performed a fixed number of times, each trial is independent, there are only two possible outcomes (success or failure), and the probability of success is constant for every trial.
- The sum of probabilities for all possible outcomes (\(k=0, 1, \ldots, n\)) in a binomial distribution is always 1.
- The shape of the binomial distribution depends on the values of \(n\) and \(p\). It is symmetric if \(p=0.5\), skewed to the right if \(p < 0.5\), and skewed to the left if \(p > 0.5\). As \(n\) increases, the binomial distribution approaches a normal distribution (Central Limit Theorem).
- The standard deviation of a binomial distribution is \(\sigma = \sqrt{npq}\).
- Real-world examples include the number of heads in a series of coin flips, the number of defective items in a production batch, or the number of successful treatments in a clinical trial.
Understanding the relationship between mean, variance, and the parameters \(n\) and \(p\) is crucial for solving problems involving binomial distributions.
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Important Questions from Binomial Distribution
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For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
Indicate the correct answer for the combination from the following regarding the conditions for the applicability of a binominal distribution:
(a) There are n independent trials
(b) Each trial has only two possible outcomes
(c) The probabilities of two outcomes do not remain constant
(d) The trials are independent
Which of the following options is correct?
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
