Question

Mean of a binomial distribution 12 and variance is 8. The probability of having at least one success is?

• A. $1 - \left(\frac{2}{3}\right)^{36}$
• B. $1 - \left(\frac{1}{3}\right)^{36}$
• C. $1 - \left(\frac{2}{3}\right)^{35} \left(\frac{1}{3}\right)$
• D. $1 - \left(\frac{1}{3}\right)^{35} \left(\frac{2}{3}\right)$

Hint:

• For $B(n,p)$: Mean $\mu = np$, Variance $\sigma^2 = npq$.
• $q = \frac{npq}{np} = \frac{8}{12} = \frac{2}{3}$ (probability of failure).
• $p = 1-q = \frac{1}{3}$ (probability of success).
• $n = \frac{np}{p} = \frac{12}{1/3} = 36$.
• $P(X \ge 1) = 1 - P(X=0)$.
• $P(X=0) = q^n = (\frac{2}{3})^{36}$.
• So $P(X \ge 1) = 1 - (\frac{2}{3})^{36}$. This is option (b).
• The question key indicates (a). This suggests an error in the key, as option (a) leads to a contradiction with the given variance if $1/3$ is interpreted as $q$. If $1/3$ in option (a) is $p$, then $P(X=0)=q^n \neq p^n$.

Answer 1

The Correct Option is A

In a binomial distribution, the mean ($\mu$) and variance ($\sigma^2$) are given by the following formulas:

\[ \mu = n \cdot p \] \[ \sigma^2 = n \cdot p \cdot (1 - p) \] where: - $n$ is the number of trials, - $p$ is the probability of success on a single trial. We are given the following information: - The mean $\mu = 12$, - The variance $\sigma^2 = 8$. 
Step 1: Use the mean to find $p$ From the formula for the mean: \[ n \cdot p = 12 \] This gives us: \[ p = \frac{12}{n} \] 
Step 2: Use the variance to find $n$ and $p$ From the formula for the variance: \[ n \cdot p \cdot (1 - p) = 8 \] Substitute $p = \frac{12}{n}$ into this equation: \[ n \cdot \frac{12}{n} \cdot \left( 1 - \frac{12}{n} \right) = 8 \] Simplifying: \[ 12 \cdot \left( 1 - \frac{12}{n} \right) = 8 \] \[ 12 - \frac{144}{n} = 8 \] \[ \frac{144}{n} = 4 \] \[ n = 36 \] 
Step 3: Calculate the probability of at least one success Now that we know $n = 36$, we can substitute this value into the equation for $p$: \[ p = \frac{12}{36} = \frac{1}{3} \] The probability of having at least one success is the complement of the probability of having zero successes. The probability of zero successes in a binomial distribution is given by: \[ P(X = 0) = (1 - p)^n = \left( 1 - \frac{1}{3} \right)^{36} = \left( \frac{2}{3} \right)^{36} \] Therefore, the probability of having at least one success is: \[ P(X \geq 1) = 1 - P(X = 0) = 1 - \left( \frac{2}{3} \right)^{36} \] This is the probability of having at least one success. 

Final Answer: The probability of having at least one success is \( 1 - \left( \frac{2}{3} \right)^{36} \).