Question

A coin is tossed three times. Find the probability of the following events:
(1) A: getting at least two heads
(2) B: getting exactly two heads
(3) C: getting at most one head

Answer 1

Sample space: $\left\{\begin{array}{c}HHH,HTH,THH,TTH\\ HHT,HTT,THT,TTT\end{array}\right\}$

Total number of possible outcomes$=8$

$\left(1\right)$ A getting at least two heads

$P\left(A\right)=P($getting two heads$)+$P$($getting $3$ heads$)$

$=\frac{3}{8}+\frac{1}{8}$

(using formula$=$P(event)$=\frac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}}$

$P\left(A\right)=\frac{4}{8}=\frac{1}{2}$

$\therefore P\left(A\right)=\frac{1}{2}$

$\left(2\right)$ B: getting exactly two heads

$P\left(B\right)=3/8$

$\left(3\right)$ C: getting almost one head

$P\left(C\right)=P($getting no head$)+$P$($getting $1$ head$)$

$=\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}$

$P\left(C\right)=\frac{1}{2}$.