# Three coins are tossed once.Find the probability of getting no tail

$\begin{array}{1 1}(A)\;\large\frac{1}{8}\\(B)\;\large\frac{1}{6}\\(C)\;\large\frac{1}{5}\\(D)\;\large\frac{1}{9}\end{array}$

Step 1:
If three coins are tossed,then the possible outcomes =$2^3=8$
$\therefore$ The sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$
Step 2:
Let E be the event of getting no tail
$\therefore n(E)=1$
$\therefore$ Probability of getting no tails=$\large\frac{n(E)}{n(S)}=\frac{1}{8}$
Hence (A) is the correct answer.