Question

A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is
(a) $\frac{{}^{20}C_{10}\times 5^6}{6^{20}}$

(b) $\frac{120\times 5^7}{6^{10}}$

(c) $\frac{84\times 5^6}{6^{10}}$

(d) None of these

Answer 1

(c) $\frac{84\times 5^6}{6^{10}}$


$$\begin{gathered} \mathrm{Let}\mathit{}p\mathit{}\mathrm{be}\mathrm{the}\mathrm{probabilty}\mathrm{of}\mathrm{obtaining}\mathrm{a}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die}.\mathrm{Then}\mathit{,} \\ p\mathit{}=\frac{1}{6}\mathrm{and}q\mathit{}=1-\frac{1}{6}=\frac{5}{6} \\ \mathrm{Obtaining}\text{a}\mathrm{fourth}\mathrm{six}\mathrm{in}\mathrm{the}\mathrm{tenth}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die}\mathrm{means}\mathrm{that}\mathrm{in}\text{the}\mathrm{first}\mathrm{nine}\mathrm{throws} \\ \mathrm{there}\mathrm{are}3\mathrm{sixes}\mathrm{and}\mathrm{the}\mathrm{fourth}\mathrm{six}\mathrm{is}\mathrm{obtained}\mathrm{in}\mathrm{the}\mathrm{tenth}\mathrm{throw}.\mathrm{Therefore},\text{r}\mathrm{equired}\mathrm{probability} \\ =P(\mathrm{Getting}3\mathrm{sixes}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{nine}\mathrm{throws})P(\mathrm{Getting}\text{a}\mathrm{six}\mathrm{in}\mathrm{the}\mathrm{tenth}\mathrm{throw}) \\ =\left({}^{9}C_3{p}^3{q}^{9-3}\right)p \\ ={}^{9}C_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^6\times \frac{1}{6} \\ =\frac{84\mathrm{x}{5}^6}{6^{10}} \end{gathered}$$