Question

A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is

• A. $\frac{{}^{20}{\mathrm{C}}_{10}\times 5^6}{6^{20}}$
• B. $\frac{120\times 5^7}{6^{10}}$
• C. $\frac{84\times 5^6}{6^{10}}$
• D. None of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Finding the probability of the fourth six appearing on the tenth throw:

Let $\mathrm{p}$ represent the chance of getting a six on a single die roll and $q$ represent the chance of not getting a six on a single die roll. Then,

$$\begin{gathered} \mathrm{p}=\frac{1}{6}\text{and} \\ \mathrm{q}=1-\frac{1}{6} \\ =\frac{5}{6} \end{gathered}$$

We should have three sixes and six non-sixes in the first nine throws, then a six in the tenth throw; after that, it doesn't matter what face emerges.

When you get a fourth six in the tenth throw of the die, it implies you got three sixes in the first nine throws.

As a result, the needed probability:

${=}^9{C}_3{\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^6\times \frac{1}{6}\times 1\times 1\times 1\times \dots \times 1\left(10\text{times}\right)$

$$\begin{gathered} =\frac{9!}{3!\left(6\right)!}{\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^6\times \frac{1}{6}\times 1\times 1\times 1\times \dots \times 1\left(10\text{times}\right)\left[\because {}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =\frac{9\times 8\times 7}{3\times 2}{\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^6\times \frac{1}{6}\times 1\times 1\times 1\times \dots \times 1\left(10\text{times}\right) \end{gathered}$$

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

Hence, the correct option is C.