Q. An unbiased coin is tossed to get 2 points for turning up a head and one point for the tail. If three unbiased coins are tossed simultaneously, then the probability of getting a total of odd number of points
A) $\frac{1}{2}$
B) $\frac{1}{4}$
C) $\frac{1}{8}$
D) $\frac{3}{8}$
Explanation
(a)
We are getting a odd number of points, if it will comes (two heads, one tail and three tails)
$\because\ P(H) = P(T) = \frac{1}{2}$
$\therefore$Required probability=Probability of getting two heads and one tail + Probability of all three tails
$= \ ^{3}C_{2}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{1} + \left( \frac{1}{2} \right)^{3}$
$= 3\left( \frac{1}{2} \right)^{3} + \left( \frac{1}{2} \right)^{3}$
$= \frac{3}{8} + \frac{1}{8} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.