A coin is tossed twice, what is the probability that atleast one tail occurs?

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Solution

Here $S = {H H, H T, T H, T T}$
$∴$ Number of possible outcomes n (S) =4
Let E be the event of getting at least one tail

$∴$ n (E) = 3
$∴$ Probability of getting at least one tail
$P (E) = \frac{n (E)}{n (S)} = \frac{3}{4}$

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A coin is tossed twice, what is the probability that atleast one tail occurs?

Here S={HH,HT,TH,TT} ∴ Number of possible outcomes n (S) =4 Let E be the event of getting at least one tail ∴ n (E) = 3 ∴ Probability of getting at least one tail P(E)=n(E)n(S)=34

Here $S = {H H, H T, T H, T T}$
$∴$ Number of possible outcomes n (S) =4
Let E be the event of getting at least one tail

$∴$ n (E) = 3
$∴$ Probability of getting at least one tail
$P (E) = \frac{n (E)}{n (S)} = \frac{3}{4}$