Question

If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

Answer 1

Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.

Then, X follows a binomial distribution with n =6;
$$\begin{gathered} p\mathit{}=\mathrm{of}\mathrm{getting}5\mathrm{or}6=\frac{1}{6}+\frac{1}{6}=\frac{1}{3};q=1-p=\frac{2}{3}; \\ P(X=r)={}^{6}C_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{6-r} \\ P(X\ge 4)=P(X=4)+P(X=5)+P(X=6) \\ ={}^{6}C_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^{\mathit{6}\mathit{-}\mathit{4}}+{}^{6}C_5{\left(\frac{1}{3}\right)}^5{\left(\frac{2}{3}\right)}^{6-5}+{}^{6}C_6{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^{6-6} \\ =\frac{1}{3^6}(60+12+1) \\ =\frac{73}{729} \end{gathered}$$