Browse Questions

# Three coins are tossed once.Find the probability of getting atleast 2 heads

$\begin{array}{1 1}(A)\;\large\frac{1}{2}\\(B)\;\large\frac{1}{4}\\(C)\;\large\frac{1}{6}\\(D)\;\large\frac{1}{5}\end{array}$

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 "atleast 2 heads"
$\therefore n(E)=$[2 heads + 3 heads]
$\Rightarrow 3+1=4$
Step 3:
$\therefore$ Probability of getting atleast 2 heads
$P=\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{4}{8}$
$\Rightarrow \large\frac{1}{2}$
Hence (A) is the correct answer.