Question

If a dice is thrown $7$ times, then the probability of obtaining $5$ exactly $4$ times is

• A. ${}^{7}C_4{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^3$
• B. ${}^{7}C_4{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$
• C. ${\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^3$
• D. ${\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$

Answer 1

The correct option is A

${}^{7}C_4{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^3$

Explanation for the correct option:

Step 1. Find the required probability:

Given, $n=7$, $x=4$

Now, Probability of getting a $5$ on a die , $p=\frac{1}{6}$

probability of not getting a $5$ on a die, $q=1-p$

$$\begin{gathered} =1-\frac{1}{6} \\ =\frac{5}{6} \end{gathered}$$

Step 2. Probability of getting $5$ exactly four times:

$\begin{array}{rcl}P(X& =& 4)={}^{7}C_4{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^{7-4}\end{array}$ ; $\left[\because P(X=x)={}^{n}C_x{p}^x{q}^{n-x}\right]$

$\begin{array}{rcl}& =& {}^{\mathbf{7}}\mathit{C}_{\mathbf{4}}{\left(\frac{1}{6}\right)}^{\mathbf{4}}{\left(\frac{5}{6}\right)}^{\mathbf{3}}\end{array}$

Hence, Option ‘A’ is Correct.