Question

If a dice is thrown twice, the probability of occurrence of $4$ atleast once, is

• A. $\frac{11}{36}$
• B. $\frac{7}{12}$
• C. $\frac{35}{36}$
• D. None of these

Answer 1

Answer: (A)

$\frac{11}{36}$

Probability of occurence of $4=\frac{1}{6}$
Probability of non-occurrence of $4=\frac{5}{6}$
$\therefore$ Required probability
$={{}^{2}C}_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)+{{}^{2}C}_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^2$
$=2.\frac{5}{36}+1.\frac{1}{36}=\frac{10}{36}+\frac{1}{36}=\frac{11}{36}$