Question

A fair dice is thrown upto $20$ times. The probability that on the $10$th throw, the fourth six appears is -

• A. $\frac{84\times 5^6}{6^{10}}$
• B. $\frac{112\times 5^6}{6^{10}}$
• C. $\frac{84\times 5^6}{6^{20}}$
• D. $\frac{84\times 5^6}{6^9}$

Answer 1

The correct option is A $\frac{84\times 5^6}{6^{10}}$

Since, the fourth six appears on the tenth throw.
Therefore $3$ six appear in first nine throws and another six appears on the tenth throw.
Required probability is
$={}^9{C}_3\times {\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^6\times \left(\frac{1}{6}\right)$

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