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

Revision Table: Binomial Distribution Concepts

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.