Question

If the mean and variance of a binomial distribution are \(\frac{10}{3}\) and \(\frac{10}{9}\) respectively, then the probability of having at least one success is

• A. \(\frac{242}{243}\)
• B. \(\frac{32}{243}\)
• C. \(\frac{31}{243}\)
• D. \(\frac{211}{243}\)

Hint:

For binomial distribution questions asking probability of “at least one”, always use complementary probability \(1-P(X=0)\). It saves time and avoids unnecessary calculations of multiple terms.

Answer 1

The Correct Option is A

Concept: A binomial distribution is used when an experiment consists of a fixed number of independent trials, where each trial has only two possible outcomes: success or failure. If a random variable follows binomial distribution, then \[ X\sim B(n,p) \] where
• \(n\) = number of trials
• \(p\) = probability of success
• \(q=1-p\) = probability of failure The two most important formulas are: Mean: \[ \mu=np \] Variance: \[ \sigma^2=npq \] Probability of exactly \(r\) successes is \[ P(X=r)=^nC_r p^r q^{\,n-r} \] If we need probability of at least one success, then instead of calculating many probabilities separately, we use complementary probability. That formula is \[ P(X\geq1)=1-P(X=0) \] This reduces calculation significantly.

Step 1: Write the given information from the question.
We are given: Mean \[ np=\frac{10}{3} \] Variance \[ npq=\frac{10}{9} \] We must find probability of at least one success. That means we eventually need \[ P(X\geq1) \]

Step 2: Determine value of q by dividing variance by mean.
We know \[ np=\frac{10}{3} \] and \[ npq=\frac{10}{9} \] Dividing second equation by first equation gives \[ q=\frac{\frac{10}{9}}{\frac{10}{3}} \] Now simplify carefully \[ q=\frac{10}{9}\times\frac{3}{10} \] Canceling common factor 10 \[ q=\frac{3}{9} \] \[ q=\frac13 \] Thus probability of failure becomes \[ q=\frac13 \]

Step 3: Determine probability of success p.
We know relation \[ p+q=1 \] Since \[ q=\frac13 \] therefore \[ p=1-\frac13 \] \[ p=\frac23 \] Thus success probability is \[ p=\frac23 \]

Step 4: Determine number of trials n.
Using mean formula \[ np=\frac{10}{3} \] Substitute value of p \[ n\left(\frac23\right)=\frac{10}{3} \] Multiply both sides by 3 \[ 2n=10 \] Divide by 2 \[ n=5 \] Hence total number of trials is \[ n=5 \]

Step 5: Apply formula for probability of at least one success.
We need \[ P(X\geq1) \] Instead of finding probability of one success, two successes, three successes and so on, we use complementary probability. \[ P(X\geq1)=1-P(X=0) \] Probability of zero success means all trials fail. Thus \[ P(X=0)=q^n \] Substitute values \[ P(X=0)=\left(\frac13\right)^5 \] \[ =\frac1{243} \] Therefore \[ P(X\geq1)=1-\frac1{243} \] \[ =\frac{243-1}{243} \] \[ =\frac{242}{243} \]

Step 6: Write final answer.
Thus required probability becomes \[ \boxed{\frac{242}{243}} \] This matches option (1).