Question

I have $3$ normal disc, one red, one blue and one green and I roll all three simultaneously. Let $P$ be the porbability that the sum of the numbers on the red and blue dice is equal to the number on the green die. If $P$ is the written in lowest terms as $a/b$ then the value of $(a+b)$ equals-

• A. $79$
• B. $77$
• C. $61$
• D. $57$

Answer 1

The correct option is B $77$

R B G

Total cases $\to$ $6\times 6\times 6=216$

Favourable cases

(i) If green $=2$, B$=1$, R$=1$

(ii)$G=3$, $\Rightarrow B=1,R=2$

$B=2,R=1$

(iii)$G=4\Rightarrow B=1,R=3$

$B=2,R=2$

$B=3,R=1$

(iv) $G=5$ $\to$ $B=1,R=4$

$B=2,R=3$

$B=3,R=2$

$B=4,R=1$

(v)$G=6$ $\Rightarrow$ $B=1,R=5$

$B=2,R=4$

$B=3,R=3$

$B=4,R=2$

$B=5,R=1$

Total favourable cases $=15$

$P=\frac{15}{216}=\frac{5}{72}=\frac{a}{b}$

$a+b=77$.