Q. A coin is tossed 10 times. The probability of getting exactly six heads, is
A) $\frac{512}{513}$
B) $\frac{105}{512}$
C) $\frac{100}{153}$
D) $\ ^{10}C_{6}$
Explanation
(b)
$\therefore P = P$(getting a head)$= \frac{1}{2}$
$\therefore\ q = \frac{1}{2}$
$\therefore$ Required probability
$= {\ ^{10}C}_{6}\left( \frac{1}{2} \right)^{6}\left( \frac{1}{2} \right)^{4}$
$= \frac{10!}{6!4!} \times \frac{1}{2^{10}} = \frac{105}{512}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.