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. $\frac{\left(120\times 5^7\right)}{6^{10}}$
• C. $\frac{84\times 5^6}{6^{10}}$
• D. None of these

Answer 1

The correct option is C

$\frac{84\times 5^6}{6^{10}}$

Find the probability that on the tenth throw the fourth six appears

To have the fourth six on the tenth throw compulsory condition is

While throwing the first $9$ time we should have $3$ sixes and the other six times there can be any number

Probability of getting six $\text{'}\mathrm{p}\text{'}=\frac{1}{6}$

Probability of getting other numbers $\text{'}\mathrm{q}\text{'}=\frac{5}{6}$

Possibility to fill the first three places $={}^9{}^{C}3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{9-3}\mathbf{}\left[Apply{}^n{}^{C}r{p}^r{q}^{n-r}\right]$

Possibility to get six on the tenth throw $=\frac{1}{6}$

Required probability $\begin{array}{rcl}& =& {}^9{}^{C}3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{9-3}\times \frac{1}{6}\end{array}$

$\begin{array}{rcl}& =& \frac{9!}{\left(9-3\right)!3!}{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^6\times \frac{1}{6}\\ & =& \frac{9\times 8\times 7}{3\times 2\times 1}\times \frac{5^6}{6^{10}}\\ & =& \frac{84\times 5^6}{6^{10}}\end{array}$

Hence, Option(C) is the correct answer.