Question

$A$ and $B$ each throw a dice. The probability that "$B$" throw is not smaller than "$A$" throw, is

• A. $\frac{1}{36}$
• B. $\frac{5}{36}$
• C. $\frac{7}{12}$
• D. $\frac{5}{12}$

Answer 1

The correct option is C $\frac{7}{12}$

Each throw a die $n\left(S\right)=6\times 6=36$

As $B$ throw is not smaller than $A$ throw which means the throw of $B$ is always greater than equal the throw of $A$.

$\therefore$ Total favourable cases $=1+2+3+...+6=21$

$\therefore$ Required probability $=\frac{21}{36}=\frac{7}{12}$