Question

An experiment succeeds thrice as often as it fails. Find the probability that in the next live trials, there will be at least 3 successes.

Answer 1

Let the probability of success be p and that of failure be q.

Therefore, p=3q. Since p+q=1 so, $3q+q=1\Rightarrow q=\frac{1}{4},p=\frac{3}{4}$

$\therefore P\left(atleast3successes\right)=P(r\ge 3)=P\left(3\right)+P\left(4\right)+P\left(5\right)$

Using $P\left(r\right){=}^n{C}_r{p}^4{q}^{n-r}$, where n=5

i.e., $P(r\ge 3){=}^5{C}_3{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^{5-3}{+}^5{C}_4{\left(\frac{3}{4}\right)}^4{\left(\frac{1}{4}\right)}^{5-4}{+}^5{C}_5{\left(\frac{3}{4}\right)}^5{\left(\frac{1}{4}\right)}^{5-5}$

$\Rightarrow =10\left(\frac{27}{1024}\right)+5\left(\frac{81}{1024}\right)+1\left(\frac{243}{1024}\right)$

So, required probability $=\frac{918}{1024}=\frac{459}{512}$.