Question

An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.

Answer 1

Let X denote the number of successes in 6 trials.

$$\begin{gathered} \text{It is}\mathrm{given}\mathrm{that}\mathrm{successes}\mathrm{are}\mathrm{twice}\text{the}\mathrm{failure}s. \\ \Rightarrow p=2q \\ p+q=1 \\ \Rightarrow 3q=1 \\ \Rightarrow q=\frac{1}{3} \\ \therefore p=1-\frac{1}{3}=\frac{2}{3} \\ n\mathit{}=6 \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \end{gathered}$$

$$\begin{gathered} P(X=r)={}^{6}C_r{\left(\frac{2}{3}\right)}^r{\left(\frac{1}{3}\right)}^{6-r},r=0,1,2.....6 \\ P(\mathrm{atleast}4\mathrm{successes})=\mathrm{P}(\mathrm{X}\ge 4) \\ =\mathit{}P(X=4)+P(X=5)+P(X=6) \\ {}^{6}C_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^{6-4}+{}^{6}C_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^{6-5}+{}^{6}C_6{\left(\frac{2}{3}\right)}^6{\left(\frac{1}{3}\right)}^{6-6} \\ =\frac{15\left(2^4\right)+6\left(32\right)+64}{3^6} \\ =\frac{240+192+64}{729} \\ =\frac{496}{729} \end{gathered}$$