Answer
(a) The probability of the event that both coins show the same face is 0.5, as there are two favorable outcomes $(HH;TT)$ out of the sample set:
$P(A)=\frac{2}{4}=0.5$
(b) The probability of the event that at least one coin is a head is 0.75, as there are three favorable outcomes $(HH;HT;TH)$ out of the sample set:
$P(B)=\frac{3}{4}=0.75$
Work Step by Step
The sample set is $S=${$HH;TH;HT;TT$}, therefore the total number of outcomes is 4.
The probalitity of an event can be calculated as:
$P(A)=\frac{\text{favorable outcomes out of the sample set}}{\text{total number of possible outcomes}}$
This means, that the probability of the event that both coins show the same face is 0.5, as there are two favorable outcomes $(HH;TT)$ out of the sample set:
$P(A)=\frac{2}{4}=0.5$
The probability of the event that at least one coin is a head is 0.75, as there are three favorable outcomes $(HH;HT;TH)$ out of the sample set:
$P(B)=\frac{3}{4}=0.75$
