Question

An experiment succeeds twice as often as it fails. If the probability that in the next 6 trials there will be least 4 successes is $\frac{31}{{12}^x}$, then the value of $x$ is ___.

Answer 1

Let p denote the probability of succeeding in the experiment.

It is given that $p=2q=2(1-p)\Rightarrow p=\frac{2}{3}$
Let X denote the number of successes in 6 trials.

Then, X follows binomial distribution with $n=6,p=\frac{2}{3}$ and $q=\frac{1}{3}$ such that
$p(X=r){=}^6{C}_r$ ${\left(\frac{2}{3}\right)}^r$ ${\left(\frac{1}{6}\right)}^{6-r};r=0,1,2,...,6$

Required probability $=(X\ge 4)=P(X=4)+P(X=5)+P(X=6)$

${=}^6{C}_4$ ${\left(\frac{2}{3}\right)}^4{\left(\frac{1}{2}\right)}^2+$ ${}^6{C}_5$ ${\left(\frac{2}{5}\right)}^5\left(\frac{1}{3}\right)+$ ${}^6{C}_6$ ${\left(\frac{2}{3}\right)}^6{\left(\frac{1}{3}\right)}^0$

$={\left(\frac{2}{4}\right)}^4\times \frac{31}{9}$

= $\frac{31}{{12}^2}$

Comparing this with $\frac{31}{{12}^x}$ , we get $x=2$.