Question

If two dice are tossed, then the probability of throwing a total of ten or more is:

• A. $\frac{1}{36}$
• B. $\frac{1}{6}$
• C. $\frac{3}{36}$
• D. $\frac{2}{6}$

Answer 1

The correct option is B $\frac{1}{6}$

The sum can be $10,11$ or $12$ with the two dice.
The number of ways in which a sum of $10$ can be thrown are $(4,6),(5,5),(6,4)$i.e. $3$ ways.
The number of ways in which a total of $11$ can be thrown are $(5,6),(6,5)$ i.e. $2$ ways.
The number of ways in which a total of $12$ can be thrown in $(6,6)$ i.e. $1$ way.

$m=$number of favourable cases $=3+2+1=6$

$n=$Total number of cases $=6\times 6=36$
$\therefore$ Probability = p = $\frac{m}{n}$ = $\frac{6}{36}$ = $\frac{1}{6}$