Question

A fair die is tossed eight times. The probability that a third six is observed in the eighth throw is
(a)

(b)

(c)

(d) None of these

Answer 1

(b) $\frac{{}^{7}C_2\times 5^5}{6^8}$


$$\begin{gathered} \mathrm{Let}\mathit{}p\mathit{}\mathrm{be}\mathrm{the}\mathrm{probabilty}\mathrm{of}\mathrm{obtaining}\mathrm{a}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die}.\mathrm{Then}\mathit{,} \\ p\mathit{}=\frac{1}{6}\mathrm{and}q\mathit{}=1-\frac{1}{6}=\frac{5}{6} \\ \mathrm{Obtaining}a\mathrm{third}\mathrm{six}\mathrm{in}\mathrm{the}\mathrm{eighth}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die}\mathrm{means}\mathrm{that}\mathrm{in}\mathrm{first}\mathrm{seven}\mathrm{throws} \\ \mathrm{there}\mathrm{are}2\mathrm{sixes}\mathrm{and}\mathrm{the}\mathrm{third}\mathrm{six}\mathrm{is}\mathrm{obtained}\mathrm{in}\mathrm{the}\mathrm{eighth}\mathrm{throw}.\mathrm{Therefore}, \\ \text{r}\mathrm{equired}\mathrm{probability} \\ =P(\mathrm{Getting}2\mathrm{sixes}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{seven}\mathrm{throws})P(\mathrm{Getting}\mathrm{six}\mathrm{in}\mathrm{the}\mathrm{eighth}\mathrm{throw}) \\ =\left({}^{7}C_2{p}^2{q}^{7-2}\right)p \\ ={}^{7}C_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^5\mathit{}\mathit{\times }\frac{1}{6} \\ =\frac{{}^{7}C_2\mathrm{x}{5}^5}{6^8} \end{gathered}$$