Question

Find the probability of getting $4$ exactly thrice in $7$ throws of a die

• A. ${}^7{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$
• B. ${}^7{C}_4{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^3$
• C. ${}^7{C}_4{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$
• D. ${}^7{C}_3{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^3$

Answer 1

Answer: (A), (C)

${}^7{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$
${}^7{C}_4{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$

Here $n=7$, probability of getting $4$ in one throw is $p=\frac{1}{6}\Rightarrow q=1-p=\frac{5}{6}$
Thus probability of getting $4$ exactly thrice in $7$ throw is
$=P(X=3){=}^7{C}_3{p}^3{q}^4{=}^7{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4{=}^7{C}_4{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^4$