Question

A pair of dice is thrown 4 times, then the probability of getting doublets at least twice is

• A. $\frac{19}{44}$
• B. $\frac{21}{44}$
• C. $\frac{31}{44}$
• D. $\frac{39}{44}$

Answer 1

Answer: (A)

$\frac{19}{44}$

$n\left(S\right)=6^2=36$
$p=$ probability of getting a doublet on the pair of dice
$\therefore P=\frac{6}{36}=\frac{1}{6},q=1-p=1-\frac{1}{6}=\frac{5}{6},n=4$
$\therefore P(X=r){=}^4{C}_r{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{4-r}$, $r=0,1,2,3,4$
$\therefore P(X=r\ge 2)=1-P(X=r<2)$ $=1-\left\{P\right(X=r=0)+P(X=r=1\left)\right\}$ $=1-\left\{{}^4{C}_0{\left(\frac{1}{6}\right)}^0{\left(\frac{5}{6}\right)}^4{+}^4{C}_1{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3\right\}$
$=1-(\frac{5}{6}+\frac{2}{3}){\left(\frac{5}{6}\right)}^3=1-\frac{375}{432}=\frac{19}{44}$
Hence choice (a) is correct answer.