Question

In a trial, the probability of success is twice the probability of failure. In six trials, the probability of at least four successes will be

• A. $\frac{496}{729}$
• B. $\frac{400}{729}$
• C. $\frac{500}{729}$
• D. $\frac{600}{729}$

Hint:

For “at least”, sum from required value to $n$ or use complement.

Answer 1

The Correct Option is A

Concept: Binomial probability: $P(X=k) = \binom{n}{k}p^k q^{n-k}$

Step 1: Find probabilities.
Let failure = $q$, success = $p = 2q$ \[ p+q=1 \Rightarrow 2q+q=1 \Rightarrow q=\frac{1}{3},\ p=\frac{2}{3} \]

Step 2: Compute $P(X \ge 4)$.
\[ P = P(4)+P(5)+P(6) \]

Step 3: Calculate each term.
\[ P(4)=\binom{6}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2 \] \[ P(5)=\binom{6}{5}\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right) \] \[ P(6)=\binom{6}{6}\left(\frac{2}{3}\right)^6 \]

Step 4: Add results.
\[ = \frac{240+192+64}{729} = \frac{496}{729} \] Conclusion:
Answer = $\frac{496}{729}$