Answer:
(a) See attachment for tree diagram
(b) 24 possible outcomes
Step-by-step explanation:
Given
[tex]Urn\ 1 = \{B_1, R_1, R_2, R_3\}[/tex]
[tex]Urn\ 2 = \{R_4, R_5, B_2, B_3\}[/tex]
Solving (a): A possibility tree
If urn 1 is selected, the following selection exists:
[tex]B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2][/tex]
If urn 2 is selected, the following selection exists:
[tex]B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4][/tex]
See attachment for possibility tree
Solving (b): The total number of outcome
For urn 1
There are 4 balls in urn 1
[tex]n = \{B_1,R_1,R_2,R_3\}[/tex]
Each of the balls has 3 subsets. i.e.
[tex]B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2][/tex]
So, the selection is:
[tex]Urn\ 1 = 4 * 3[/tex]
[tex]Urn\ 1 = 12[/tex]
For urn 2
There are 4 balls in urn 2
[tex]n = \{B_2,B_3,R_4,R_5\}[/tex]
Each of the balls has 3 subsets. i.e.
[tex]B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4][/tex]
So, the selection is:
[tex]Urn\ 2 = 4 * 3[/tex]
[tex]Urn\ 2 = 12[/tex]
Total number of outcomes is:
[tex]Total = Urn\ 1 + Urn\ 2[/tex]
[tex]Total = 12 + 12[/tex]
[tex]Total = 24[/tex]
