Question

An experiment succeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Answer 1

If p is the probability of a success and q, that of a failure, then p+q=1 and p=2q

$2q+q=1\Rightarrow q=\frac{1}{3}andhence,p=\frac{2}{3}$

Let X be the random variable that represents the number of successes in six trials.

Also, n=6

By Binomial distribution, we obtain

$P(X=r){=}^n{C}_r{p}^{r}qn-r$

${=}^6{C}_r{\left(\frac{2}{3}\right)}^r{\left(\frac{1}{3}\right)}^{6-r}$

P(atleast 4 success in 6 trials) = $P(X\ge 4)$ = P(4)+P(5)+P(6)

${=}^6{C}_4{p}^4{q}^2+{=}^6{C}_5{p}^5{q}^1+{=}^6{C}_6{p}^6{q}^0=p^4{}^6{C}_2{q}^2{+}^6{C}_{1}pq{+}^6{C}_0{p}^2\left({\because }^n{C}_r{=}^n{C}_{n-r}\right)={\left(\frac{2}{3}\right)}^4[\frac{6\times 5}{1\times 2}{\left(\frac{1}{3}\right)}^2+\frac{6}{1}.\frac{2}{3}.\frac{1}{3}+{\left(\frac{2}{3}\right)}^2]={\left(\frac{2}{3}\right)}^4[\frac{15}{9}+\frac{12}{9}+\frac{4}{9}]=\frac{31}{9}{\left(\frac{2}{3}\right)}^4$