Question

Three fair dice are thrown. The probability of getting a sum $6$ or less on the three dice is

• A. $\frac{7}{12}$
• B. $\frac{5}{54}$
• C. $\frac{19}{216}$
• D. $\frac{11}{108}$

Answer 1

The correct option is B $\frac{5}{54}$

Sum can be six in following ways

Die 1 Die 2 Die 3

$1111222334$ $1234123121$ $4321321211$

No. of ways $=10$

Total no. of possible outcomes $=6\times 6\times 6=216$

$\therefore$ Probability $=\frac{10}{216}=\frac{5}{108}$

Sum can be $5$ in following ways

$111223$ $123121$ $321211$ $P=\frac{6}{216}=\frac{3}{108}$

Sum can be $4$ in following ways

$112$ $121$ $211$ $P=\frac{3}{216}$

Sum can be $3$ in following way :

$1$ $1$ $1$ $P=\frac{1}{216}$

$\therefore$ Total probability $=\frac{5}{108}+\frac{3}{108}+\frac{3}{216}+\frac{1}{216}=\frac{5+3+2}{108}=\frac{5}{54}$ [B]