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Home  >>  CBSE XI  >>  Math  >>  Probability
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A coin is tossed twice,what is the probability that atleast one tail occurs

$\begin{array}{1 1}(A)\;\large\frac{3}{4}\\(B)\;\large\frac{4}{5}\\(C)\;\large\frac{3}{5}\\(D)\;\large\frac{2}{3}\end{array} $

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1 Answer

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  • Required probability =$\large\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$
Step 1:
Given a coin is tossed twice.
The sample space $S=\{HH,HT,TH,TT\}$
Let E be the event of getting atleast one tail
$\therefore$ E={HT,TH,TT}
Step 2:
$\therefore$ Required probability =$\large\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{3}{4}$
$\therefore$ The Answer is $\large\frac{3}{4}$
Hence (A) is the correct answer.
answered Jun 30, 2014 by sreemathi.v
 

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