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

Answer 1

The correct option is A

$\frac{496}{729}$

Explanation for the correct option:

Find the probability:

Let $p$ is the probability of success and $q$ be the probability of failure.

According to the question, the probability of success is twice the probability of failure. Thus,

$p=2q$

As we know, $p+q=1$

So, $2q+q=1$

$$\begin{gathered} 3q=1 \\ q=\frac{1}{3}andp=\frac{2}{3} \end{gathered}$$

Hence, the required probability $=$ P(Four success) $+$ P(Five success) $+$ P(Six success)

$$\begin{gathered} {=}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{C}_5{\left(\frac{2}{3}\right)}^5\left(\frac{1}{3}\right){+}^6{C}_6{\left(\frac{2}{3}\right)}^6 \\ =\frac{6!}{4!2!}{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2+\frac{6!}{5!1!}{\left(\frac{2}{3}\right)}^5\left(\frac{1}{3}\right)+{\left(\frac{2}{3}\right)}^6 \\ =15{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2+6{\left(\frac{2}{3}\right)}^5\left(\frac{1}{3}\right)+{\left(\frac{2}{3}\right)}^6 \\ =\frac{496}{729} \end{gathered}$$

Hence, the correct option is A.