Q. A die has four blank faces and two faces marked 3. The change of getting a total of 12 in 5 throws is
A) $\ ^{5}C_{4}\left( \frac{1}{3} \right)^{4}\left( \frac{2}{3} \right)$
B) $\ ^{5}C_{4}\left( \frac{1}{3} \right)^{\ }\left( \frac{2}{3} \right)^{4}$
C) $\ ^{5}C_{4}{\left( \frac{1}{6} \right)^{\ }}^{5}$
D) $\ ^{5}C_{4}\left( \frac{1}{6} \right)^{4}\left( \frac{5}{6} \right)$
Explanation
(a)
Probability of success, $P = \frac{2}{6} = \frac{1}{3}$ and probability of failure, $q = \frac{2}{3}$
$\therefore$ Required probability$= \ ^{5}C_{4}\left( \frac{1}{3} \right)^{4}\left( \frac{2}{3} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.