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A coin is tossed twice. Then, the probability that atleast one tail occurs is
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Verified Answer
The correct answer is:
$\frac{3}{4}$
The sample space is $\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$
Let $\mathrm{E}$ be the event of getting atleast one tail
$\begin{aligned}
& \therefore \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}, \mathrm{TT}\} \\
& \therefore \text { Required probability } \mathrm{p} \\
& =\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{3}{4}
\end{aligned}$
Let $\mathrm{E}$ be the event of getting atleast one tail
$\begin{aligned}
& \therefore \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}, \mathrm{TT}\} \\
& \therefore \text { Required probability } \mathrm{p} \\
& =\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{3}{4}
\end{aligned}$
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