Answer:
(a)
[tex]S = \{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}[/tex]
(b)
[tex]P(A) = \frac{1}{6}[/tex]
Step-by-step explanation:
Given
The above sample space
A = Event of an occurrence of 3 or 4 then a head
(a): List out the sample space:
The rolls of the dice come first, followed by the toss of the coin.
So, the sample space are:
[tex]S = \{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}[/tex]
The number of the sample space is:
[tex]n(S) = 12[/tex]
(b): Find P(A)
First, we list out the outcomes of event A
[tex]A = \{(3,H),(4,H)\}[/tex]
The number of outcome is:
[tex]n(A) = 2[/tex]
The probability of A is then calculated as:
[tex]P(A) = \frac{n(A)}{n(S)}[/tex]
[tex]P(A) = \frac{2}{12}[/tex]
Simplify fraction
[tex]P(A) = \frac{1}{6}[/tex]
