Probability

Three coins are tossed once. Find the probability of getting

1. exactly two heads;
2. at least two heads;
3. at most two heads.

When $3$ coins are tossed, number of elements in sample space $S = n\left(S\right) = 2^3 = 8$

1. Let $A$ be the event of getting exactly $2$ heads
   
   Then, $A = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right)\right\}$
   
   $\therefore \;$ Number of elements in $A = n\left(A\right) = 3$
   
   $\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{3}{8}$
   
   
2. Let $B$ be the event of getting at least $2$ heads
   
   Then, $B = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right), \left(H,H,H\right)\right\}$
   
   $\therefore \;$ Number of elements in $B = n\left(B\right) = 4$
   
   $\therefore \;$ Probability of event $B = P\left(B\right) = \dfrac{n\left(B\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$
   
   
3. Let $C$ be the event of getting at most $2$ heads
   
   Then, $C = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right), \left(H,T,T\right), \right.$
   $\hspace{4cm}$ $\left. \left(T,H,T\right), \left(T,T,H\right), \left(T,T,T\right)\right\}$
   
   $\therefore \;$ Number of elements in $C = n\left(C\right) = 7$
   
   $\therefore \;$ Probability of event $C = P\left(C\right) = \dfrac{n\left(C\right)}{n \left(S\right)} = \dfrac{7}{8}$
   
   ALTERNATIVELY
   
   $\begin{aligned} P\left(\text{getting at most 2 heads}\right) & = 1 - P \left(\text{getting 3 heads}\right) \\\\ & = 1 - \dfrac{1}{8} \\\\ & = \dfrac{7}{8} \end{aligned}$