Answer:
The 3rd option or 39,600 would be the correct answer.
Step-by-step explanation:
Hope this helps:)
 
        
             
        
        
        
Answer:
(a) See attachment for tree diagram
(b) 24 possible outcomes
Step-by-step explanation:
Given


Solving (a): A possibility tree
If urn 1 is selected, the following selection exists:
![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]](https://tex.z-dn.net/?f=B_1%20%5Cto%20%5BR_1%2C%20R_2%2C%20R_3%5D%3B%20R_1%20%5Cto%20%5BB_1%2C%20R_2%2C%20R_3%5D%3B%20R_2%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_3%5D%3B%20R_3%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_2%5D)
If urn 2 is selected, the following selection exists:
![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]](https://tex.z-dn.net/?f=B_2%20%5Cto%20%5BB_3%2C%20R_4%2C%20R_5%5D%3B%20B_3%20%5Cto%20%5BB_2%2C%20R_4%2C%20R_5%5D%3B%20R_4%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_5%5D%3B%20R_5%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_4%5D)
<em>See attachment for possibility tree</em>
Solving (b): The total number of outcome
<u>For urn 1</u>
There are 4 balls in urn 1

Each of the balls has 3 subsets. i.e.
![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]](https://tex.z-dn.net/?f=B_1%20%5Cto%20%5BR_1%2C%20R_2%2C%20R_3%5D%3B%20R_1%20%5Cto%20%5BB_1%2C%20R_2%2C%20R_3%5D%3B%20R_2%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_3%5D%3B%20R_3%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_2%5D)
So, the selection is:


<u>For urn 2</u>
There are 4 balls in urn 2

Each of the balls has 3 subsets. i.e.
![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]](https://tex.z-dn.net/?f=B_2%20%5Cto%20%5BB_3%2C%20R_4%2C%20R_5%5D%3B%20B_3%20%5Cto%20%5BB_2%2C%20R_4%2C%20R_5%5D%3B%20R_4%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_5%5D%3B%20R_5%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_4%5D)
So, the selection is:


Total number of outcomes is:



 
        
             
        
        
        
Answer:
The probability of getting heads on the toss of a coin is 0.5. If we consider all possible outcomes of the toss of two coins as shown, there is only one outcome of the four in which both coins have come up heads, so the probability of getting heads on both coins is 0.25. The second useful rule is the Sum Rule.
 
        
                    
             
        
        
        
A = 11 and b = 8 I’m pretty sure brosky