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