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

$\begin{array}{1 1}(A)\;\large\frac{1}{8}\\(B)\;\large\frac{3}{8}\\(C)\;\large\frac{1}{4}\\(D)\;\large\frac{1}{18}\end{array} $

1 Answer

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:
2 heads
Let E be the event of getting 2 heads
$\Rightarrow n(E)=3$
$\therefore$ Probability of getting 2 heads
$P=\large\frac{n(E)}{n(S)}=\frac{3}{8}$
Hence (B) is the correct answer.
answered Jul 1, 2014 by sreemathi.v
 
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