Q. The probability that in the toss of two dice, we obtain the sum 7 or 11,is
A) $\frac{1}{6}$
B) $\frac{1}{18}$
C) $\frac{2}{9}$
D) $\frac{23}{108}$
Explanation
(c)
$n(S) = 36$
Let $E =$Event of getting sum 7
$= \{(1,6),(2,5),\ (3,4),(4,3),(5,2),(6,1)\}$
$\therefore\ n(E) = 6$
$E =$Event of getting sum 11
$= \{(6,5),\ (5,6)\}$
$\therefore\ n(F) = 2$
Also $n(E \cap F) = 0$
$\therefore\ n(E \cup F) = n(E) + n(F) - n(E \cap F)$
$= 6 + 2 = 8$
$\therefore$ Required probability$= \frac{8}{36} = \frac{2}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.