An unbiased coin is tossed to get 2 points for turning up a head and one point for the tail. If three…
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 is
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{3}{8}$
Solution
We are getting a odd number of point, if it will comes (two head, one tail and three tail)
$
\because \quad P(H)=P(T)=\frac{1}{2}
$
$\therefore$ Required probability $=$ Probability of getting two heads and one tail + Probability of all three tails
$
\begin{aligned}
& ={ }^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}
\end{aligned}
$