Question

The probability of getting $5$ exactly twice in $7$ throws of a die is:

• A. $\frac{5}{12}{\left(\frac{7}{6}\right)}^5$
• B. $\frac{7}{12}{\left(\frac{5}{6}\right)}^4$
• C. $\frac{7}{12}{\left(\frac{5}{6}\right)}^5$
• D. $\frac{5}{12}{\left(\frac{7}{6}\right)}^4$

Answer 1

The correct option is C $\frac{7}{12}{\left(\frac{5}{6}\right)}^5$

Let us assume $X$ represents the number of times of getting $5$ in $7$ throws of the die.
Also, the repeated tossing of a die are the Bernoulli trials
Thus, probability of getting $5$ in a single throw, $p=\frac{1}{6}$
And, $q=1-p=1-\frac{1}{6}=\frac{5}{6}$
Clearly, we have $X$ has the binomial distribution where $n=7$ and $p=\frac{1}{6}$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$
${=}^7{C}_x{\left(\frac{5}{6}\right)}^{7-x}.{\left(\frac{1}{6}\right)}^x$
Hence, probability of getting $5$ exactly twice in a die $=P(X=2)$
${=}^7{C}_2{\left(\frac{5}{6}\right)}^5.{\left(\frac{1}{6}\right)}^2=21.{\left(\frac{5}{6}\right)}^5.\frac{1}{36}=\left(\frac{7}{12}\right){\left(\frac{5}{6}\right)}^5$