Question:
A coin is tossed twice, what is the probability that at least one tail occurs?
Solution:
When a coin is tossed twice, the sample space is given by
S = {HH, HT, TH, TT}
Let A be the event of the occurrence of at least one tail.
Accordingly, A = {HT, TH, TT}
$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}$
$=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}$
$=\frac{3}{4}$
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