When three coins are tossed simultaneously, the total number of outcomes = 23 = 8, and the sample space is given by
S = {(H,H,H),(H,H,T),(H,T,T),(H,T,H),(T,H,T),(T,T,H),(T,H,H),(T,T,T)}
Let P(A) be the probability of getting 3 heads.
The sample space of A = {(H,H,H)}
\(\therefore P(A)=\frac{1}{8}\)
Let P(B) be the probability of getting at least head.
Probability of one head = 1 – probability of no heads = 1 - 1/8 = 7/8
\(\therefore P(A)=\frac{7}{8}\)
The probability that the throw is either all heads or at least one head i.e. P(A ∪ B) = \(\frac{7}{8}\)
Now,
Tip – By conditional probability, P(A/B) = \(\frac{P(A \cup B)}{P(B)}\) where P(A/B) is the probability of occurrence of the event
A given that B has already occurred.
The probability that all coins show heads if at least one of the coins
showed a head:
