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Home  >>  CBSE XI  >>  Math  >>  Probability
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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} $

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1 Answer

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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:
Atleast 2 heads(min 2 and more heads)
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.
answered Jul 1, 2014 by sreemathi.v
 
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