A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared?

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A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared?

A coin has 2 sides and its sample space $\mathrm{S}=\{\mathrm{H}, \mathrm{T}\}$
The total number of outcomes $=2$
A coin is tossed twice.
Let $\mathrm{P}(\mathrm{A})$ be the probability of getting at most 1 tail.
The sample space of $A=\{(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H})\}$
Let $P(B)$ be the probability of getting a head.
The sample space of $B=\{H\}$
$\therefore \mathrm{P}(\mathrm{B})=\frac{1}{2}$
The probability of getting at most one tail and a head
$\begin{array}{l} \text { i. e. }(A \cap B)=\{(H, H)\} \\ \therefore P(A \cap B)=\frac{1}{3} \end{array}$
Tip – By conditional probability, $P(A / B)=\frac{P(A \cap B)}{P(B)}$ where $P(A / B)$ is the probability of occurrence of the event $A$ given that $B$ has already occurred.
The probability that both head and tail have appeared:
$\begin{array}{l} P(A / B) \\ =\frac{P(A \cap B)}{P(B)} \end{array}$
$=\frac{1 / 3}{1 / 2}$
$=\frac{2}{3}$