Let H denote the event of a head and T denote the event of a tail,

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

The problem can be solved by dividing it into 3 cases

Case 1: Head is encountered and the corresponding number on the die is Odd

Total number of sample space=(1×3)=3

We define the possible outcomes by an ordered set (x , y)

x denotes the first event the head is encountered

y denotes the second event the number on the die is Odd

Sample space S₁={(H,1),(H,3),(H,5)}

Case 2: Head is encountered and the corresponding number is even

Total number of sample space=(1×3×6)=18

We define the possible outcomes by an ordered set (x, y, z)

where

x denotes the outcomes when the coin is tossed and head is encountered

y denotes the outcomes when the die is tossed and the number encountered is even

z denotes the outcomes when the die is tossed for the second time

Sample Space

S₂={(H,2,1),(H,2,2),(H,2,3),(H,2,4),(H,2,5),(H,2,6),(H,4,1),(H,4,2),(H,4,3),(H,2,4),(H,4,5),(H,4,6), (H,6,1),(H,6,2),(H,6,3),(H,6,4),(H,6,5),(H,6,6)}

Case 3: Tail is encountered

Total number of sample space=1

Sample space S₃={(T)}

The overall sample spaces

S={(H,1),(H,3),(H,5), (H,2,1),(H,2,2),(H,2,3),(H,2,4),(H,2,5),(H,2,6),(H,4,1),(H,4,2),(H,4,3),(H,2,4),(H,4,5),(H,4,6), (H,6,1),(H,6,2),(H,6,3),(H,6,4),(H,6,5),(H,6,6),(T)}