Question

If the mean and variance of a binomial variate \(X\) are \(\frac{7}{3}\) and \(\frac{14}{9}\) respectively. Then probability that \(X\) takes value 6 or 7 is equal to (1) \(\frac{1}{729}\) (2) \(\frac{5}{729}\) (3) \(\frac{7}{729}\) (4) \(\frac{13}{729}\)

Short answer

The probability is \( \frac{5}{729} \).

Step-by-step solution

Identify the Given Values

The mean of the binomial variate is \(\frac{7}{3}\) and the variance is \(\frac{14}{9}\).

Recall the Formulas

For a binomial distribution with parameters \(n\) (number of trials) and \(p\) (probability of success), the mean \(\text{E}(X)\) and variance \( \text{Var}(X) \) are given by: \[ \text{E}(X) = np \] \[ \text{Var}(X) = np(1-p) \]

Set Up the Equations

From the given mean and variance, we get the equations: \[ np = \frac{7}{3} \] \[ np(1-p) = \frac{14}{9} \]

Solve for \p\

Divide the variance equation by the mean equation: \[ \frac{np(1-p)}{np} = \frac{\frac{14}{9}}{\frac{7}{3}} \] Simplify this to find \(p\): \[ 1 - p = \frac{14}{21} \] \[ p = \frac{1}{3} \]

Solve for \

Using the value of \(p\) in the mean equation: \[ n \times \frac{1}{3} = \frac{7}{3} \] \[ n = 7 \]

Use the Binomial Formula

The probability formula for the binomial distribution is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Calculate for \( k = 6 \): \[ P(X=6) = \binom{7}{6} (\frac{1}{3})^6 (\frac{2}{3})^1 = 7 \times \frac{1}{729} \times \frac{2}{3} = \frac{14}{2187} \] Calculate for \( k = 7 \): \[ P(X=7) = \binom{7}{7} (\frac{1}{3})^7 (\frac{2}{3})^0 = 1 \times \frac{1}{2187} \times 1 = \frac{1}{2187} \]

Find the Total Probability

Add the probabilities for \( k = 6 \) and \( k = 7 \): \[ P(X=6) + P(X=7) = \frac{14}{2187} + \frac{1}{2187} = \frac{15}{2187} = \frac{5}{729} \]

Key concepts

Mean of Binomial Distribution

In a binomial distribution, the mean gives us an idea of what value we should expect on average. For a binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success), the mean is calculated using the formula:
\[ \text{E}(X) = np \]
Here, \( n \) represents how many times an experiment is conducted, and \( p \) is the chance of succeeding in each trial. The mean tells us the expected number of successes after all trials.
In our example, the mean is given as \( \frac{7}{3} \). By setting up the equation \( np = \frac{7}{3} \), we use it to find the probability of success (\( p \)) and the number of trials (\( n \)).

Variance of Binomial Distribution

The variance measures how much the values of the binomial distribution spread out from the mean. It provides information about the distribution's consistency. The formula for variance in a binomial distribution is:
\[ \text{Var}(X) = np(1-p) \]
This formula tells us that variance depends on both the number of trials (\( n \)) and the probability of success (\( p \)).
In the given exercise, the variance is \( \frac{14}{9} \). We use this value along with the mean to solve for \( p \). By dividing the variance equation by the mean equation, we simplify and find \( p = \frac{1}{3} \). Knowing \( p \) helps us then find \( n \) using the mean equation, giving us \( n = 7 \).

Probability in Binomial Distribution

The probability of a specific outcome in a binomial distribution is computed using the binomial formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here, \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials. \( p^k \) is the probability of getting \( k \) successes, and \( (1-p)^{n-k} \) is the probability of getting the remaining \( n-k \) failures.
In the exercise, we need the probabilities for when \( X = 6 \) and \( X = 7 \).
For \( k = 6 \):
\[ P(X=6) = \binom{7}{6} (\frac{1}{3})^6 (\frac{2}{3})^1 \]
For \( k = 7 \):
\[ P(X=7) = \binom{7}{7} (\frac{1}{3})^7 (\frac{2}{3})^0 \]
By calculating these values, we obtain the probabilities for each scenario and sum them to get the total probability that \( X \) takes the value 6 or 7: \( \frac{5}{729} \).