Question

A fair die is tossed eight times.

The probability that a third six is observed on the eight throws, is

• A. $\frac{({}^{7}C_2+5^5)}{6^7}$
• B. $\frac{({}^{7}C_2\times 5^5)}{6^8}$
• C. $\frac{({}^{7}C_2+5^5)}{6^6}$
• D. None of these

Answer 1

The correct option is B

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

The explanation for the correct option:

Step1.Defined events:

Number of times a fair die $=8$

Possible outcome $=\left\{1,2,3,4,5,6\right\}$

Step2. Calculate the required probability:

Probability of getting a $6=\frac{1}{6}$

The probability of not getting a $6=\frac{5}{6}$

According to the question,

The probability of getting the third Six is observed on the eight throws

Then the two sixes must be in the first seven throws,

Probability of getting two sixes in seven throws,

$={{}^{7}C_2\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^{7-2}\mathbf{[}\mathbf{\because }\mathit{P}\mathbf{(}\mathit{x}\mathbf{=}\mathit{r}\mathbf{)}\mathbf{=}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{}{\mathbf{p}}^{\mathbf{r}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{r}}\mathbf{]}$

The probability of getting $6$ in the eighth throw is$\frac{1}{6}$.

Required probability $=$ (Probability of getting two sixes in seven throws)$\times$(The probability of getting $6$ in the eighth throw)

$={}^{7}C_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^{7-2}\times \left(\frac{1}{6}\right)$

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

Hence, Option(B) is the correct answer.