Question

Two coins are tossed simultaneously. Find the probability that either exactly two heads or atleast one tail turn up

Answer 1

Sample space $S=\{HH,HT,TH,TT\}\therefore n\left(S\right)=4$
Let, $A$ be the event of getting exactly two heads.
Then $A=\left\{HH\right\}$ and $n\left(A\right)=1\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{4}$
Let, $B$ be the event of getting atleast one tail.
Then $B=\{HT,TH,TT\}$ and $n\left(B\right)=3$
$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{3}{4}$
Now, the events $A$ and $B$ are mutually exclusive.
$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)(\because$ addition rule of probability)
$P(A\cup B)=\frac{1}{4}+\frac{3}{4}=1$
$\therefore$ the probability of getting exactly two heads or at least one tail is $1$.