Question

An unbiased die with faces $1,2,3,4,5,6$ is tossed $4$ times out of four face values obtained the probability that minimum face value is not less than $2$ and the maximum face value is not greater than $5$ is

• A. $\frac{1}{81}$
• B. $\frac{80}{81}$
• C. $\frac{64}{81}$
• D. $\frac{16}{81}$

Answer 1

The correct option is D $\frac{16}{81}$

Let the desired event be called $E$.
Hence $E$ will take $4$ values from $2,3,4,5$ with replacement.
This is because minimum value can be any of $2,3,4,5$ and maximum value can also be any of $2,3,4,5$.
so $E=$ {All same}$+${$3$ same $1$ different}$+${$2$ same $2$ different} $+$ {$2$ same $2$ other same}$+${All different}
hence $P\left(E\right)=\left(\begin{array}{c}4\\ 1\end{array}\right)\times \left({\frac{1}{6}}^4\right)+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 2\end{array}\right)\times \frac{4!}{2!}\times \left({\frac{1}{6}}^4\right)+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{4!}{3!}\times \left({\frac{1}{6}}^4\right)+\left(\begin{array}{c}4\\ 2\end{array}\right)\frac{4!}{2!\times 2!}\times \left({\frac{1}{6}}^4\right)+\left(\begin{array}{c}4\\ 4\end{array}\right)\times 4!\times \left({\frac{1}{6}}^4\right)=\left({\frac{1}{6}}^4\right)[\left(\begin{array}{c}4\\ 1\end{array}\right)+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 2\end{array}\right)\times \frac{4!}{2!}+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{4!}{3!}+\left(\begin{array}{c}4\\ 2\end{array}\right)\frac{4!}{2!\times 2!}+\left(\begin{array}{c}4\\ 4\end{array}\right)\times 4!]=\left({\frac{1}{6}}^4\right)[4+144+48+36+24]=\frac{256}{16\times 81}=\frac{16}{81}$