Question

$A$ and $B$ throw a dice. The probability that $A$'s throw is not greater that $B$'s is

• A. $\frac{5}{12}$
• B. $\frac{7}{12}$
• C. $\frac{1}{6}$
• D. $\frac{1}{2}$

Answer 1

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

If $B$ throws $1$, then $A$ can throw only $1$, if $B$ throws $2$, then $A$ can throw $1$ and $2$ and so on
$\therefore$ The required probability
$=\frac{1}{6}.\frac{1}{6}+\frac{1}{6}.\frac{2}{6}+\frac{1}{6}.\frac{3}{6}+\frac{1}{6}.\frac{4}{6}+\frac{1}{6}.\frac{5}{6}+\frac{1}{6}.\frac{6}{6}$
$=\frac{1}{6\times 6}(1+2+3+4+5+6)=\frac{6\times 7}{6\times 6\times 2}=\frac{7}{12}$