Question

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

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

Answer 1

The correct option is A $\frac{496}{729}$

Let the probability of success and failure be $p$ and $q$, respectively.
Then, $p=2q$
and $p+q=1$
$\Rightarrow 3q=1$
$\Rightarrow q=\frac{1}{3}$
Therefore, $p=1-\frac{1}{3}=\frac{2}{3}$
Required probability $={}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2+{}^6{C}_5{\left(\frac{2}{3}\right)}^6\left(\frac{1}{3}\right)+{}^6{C}_6{\left(\frac{2}{3}\right)}^6$
$=\frac{496}{729}$.