Question

A fair die is thrown 3 times . The chance that sum of three numbers appearing on the die is less than 11 , is equal to -

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{5}{8}$

Answer 1

The correct option is A $\frac{1}{2}$

From given, we have,

Sum 3:$(1,1,1)$ ==> Contributing only $1$ distinct triplet.

Sum 4: $(1,1,2)$ ==> Contributing $3$ distinct triplets

Sum 5: $(1,2,2)$and $(1,1,3)$ ==> Contributing $6$ distinct triplets

Sum 6: $(1,1,4),(1,2,3)$ and $(2,2,2)$ => Together contributing $10$distinct triplets

Sum 7: $(1,1,5),(1,2,4),(1,3,3)$and $(2,2,3)$ => Together contributing $15$ distinct triplets.

Sum 8: $(1,1,6),(1,2,5),(1,3,4),(2,3,3)$ and$(2,4,2)$==> Together contributing $21$ distinct triplets.

Sum 9: $(1,2,6),(1,3,5),(1,4,4),(2,3,4),(2,5,2)$ and $(3,3,3)$ ==> Together contributing $25$ distinct triplets.

Sum 10: $(1,3,6),(1,4,5),(2,2,6),(2,3,5),(2,4,4)$ and $(3,3,4)$==> Together contributing $27$ distinct triplets.

Therefore number of favorable cases $=1+3+6+10+15+21+25+27=108.$

Therefore, probability $=\frac{108}{216}=\frac{1}{2}$