Question

A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?

Answer 1

Let X be the number of successes in 6 throws of the two dice.

Probability of success = Probability of getting a total of 9
= Probability of getting (3,6), (4,5), (5,4), (6,3) out of 36 outcomes

$$\begin{gathered} p=\frac{4}{36}=\frac{1}{9},q=1-p\mathit{}=\frac{8}{9}\mathrm{and}n=6 \\ X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}n=6,p=\frac{1}{9}\mathrm{and}q=\frac{8}{9} \\ P(X=r)={}^{6}C_r{\left(\frac{1}{9}\right)}^r{\left(\frac{8}{9}\right)}^{6-r} \\ \mathrm{The}\mathrm{required}\mathrm{probability}=\mathrm{Prob}\text{ability}\mathrm{of}\mathrm{at}\mathrm{least}5\mathrm{successes} \\ =P(X\ge 5) \\ =P(X=5)+P(X=6) \\ ={}^{6}C_5{\left(\frac{1}{9}\right)}^5{\left(\frac{8}{9}\right)}^{6-5}+{}^{6}C_6{\left(\frac{1}{9}\right)}^6{\left(\frac{8}{9}\right)}^{6-6} \\ =\frac{6\left(8\right)+1}{9^6} \\ =\frac{49}{9^6} \end{gathered}$$