$\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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- 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.

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