Answer:
(a)
![S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7BGGG%2C%20GGB%2C%20GBG%2C%20GBB%2C%20BBG%2C%20BGB%2C%20BGG%2C%20BBB%5C%7D)
(b)
i.
![1\ girl = \{GBB, BBG, BGB\}](https://tex.z-dn.net/?f=1%5C%20girl%20%3D%20%5C%7BGBB%2C%20BBG%2C%20BGB%5C%7D)
![P(1\ girl) = 0.375](https://tex.z-dn.net/?f=P%281%5C%20girl%29%20%3D%200.375)
ii.
![Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}](https://tex.z-dn.net/?f=Atleast%5C%202%20%5C%20girls%20%3D%20%5C%7BGGG%2C%20GGB%2C%20GBG%2C%20BGG%5C%7D)
![P(Atleast\ 2 \ girls) = 0.5](https://tex.z-dn.net/?f=P%28Atleast%5C%202%20%5C%20girls%29%20%3D%200.5)
iii.
![No\ girl = \{BBB\}](https://tex.z-dn.net/?f=No%5C%20girl%20%3D%20%5C%7BBBB%5C%7D)
![P(No\ girl) = 0.125](https://tex.z-dn.net/?f=P%28No%5C%20girl%29%20%3D%200.125)
Step-by-step explanation:
Given
![Children = 3](https://tex.z-dn.net/?f=Children%20%3D%203)
![B = Boys](https://tex.z-dn.net/?f=B%20%3D%20Boys)
![G = Girls](https://tex.z-dn.net/?f=G%20%3D%20Girls)
Solving (a): List all possible elements using set-roster notation.
The possible elements are:
![S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7BGGG%2C%20GGB%2C%20GBG%2C%20GBB%2C%20BBG%2C%20BGB%2C%20BGG%2C%20BBB%5C%7D)
And the number of elements are:
![n(S) = 8](https://tex.z-dn.net/?f=n%28S%29%20%3D%208)
Solving (bi) Exactly 1 girl
From the list of possible elements, we have:
![1\ girl = \{GBB, BBG, BGB\}](https://tex.z-dn.net/?f=1%5C%20girl%20%3D%20%5C%7BGBB%2C%20BBG%2C%20BGB%5C%7D)
And the number of the list is;
![n(1\ girl) = 3](https://tex.z-dn.net/?f=n%281%5C%20girl%29%20%3D%203)
The probability is calculated as;
![P(1\ girl) = \frac{n(1\ girl)}{n(S)}](https://tex.z-dn.net/?f=P%281%5C%20girl%29%20%3D%20%5Cfrac%7Bn%281%5C%20girl%29%7D%7Bn%28S%29%7D)
![P(1\ girl) = \frac{3}{8}](https://tex.z-dn.net/?f=P%281%5C%20girl%29%20%3D%20%5Cfrac%7B3%7D%7B8%7D)
![P(1\ girl) = 0.375](https://tex.z-dn.net/?f=P%281%5C%20girl%29%20%3D%200.375)
Solving (bi) At least 2 are girls
From the list of possible elements, we have:
![Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}](https://tex.z-dn.net/?f=Atleast%5C%202%20%5C%20girls%20%3D%20%5C%7BGGG%2C%20GGB%2C%20GBG%2C%20BGG%5C%7D)
And the number of the list is;
![n(Atleast\ 2 \ girls) = 4](https://tex.z-dn.net/?f=n%28Atleast%5C%202%20%5C%20girls%29%20%3D%204)
The probability is calculated as;
![P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}](https://tex.z-dn.net/?f=P%28Atleast%5C%202%20%5C%20girls%29%20%3D%20%5Cfrac%7Bn%28Atleast%5C%202%20%5C%20girls%29%7D%7Bn%28S%29%7D)
![P(Atleast\ 2 \ girls) = \frac{4}{8}](https://tex.z-dn.net/?f=P%28Atleast%5C%202%20%5C%20girls%29%20%3D%20%5Cfrac%7B4%7D%7B8%7D)
![P(Atleast\ 2 \ girls) = 0.5](https://tex.z-dn.net/?f=P%28Atleast%5C%202%20%5C%20girls%29%20%3D%200.5)
Solving (biii) No girl
From the list of possible elements, we have:
![No\ girl = \{BBB\}](https://tex.z-dn.net/?f=No%5C%20girl%20%3D%20%5C%7BBBB%5C%7D)
And the number of the list is;
![n(No\ girl) = 1](https://tex.z-dn.net/?f=n%28No%5C%20girl%29%20%3D%201)
The probability is calculated as;
![P(No\ girl) = \frac{n(No\ girl)}{n(S)}](https://tex.z-dn.net/?f=P%28No%5C%20girl%29%20%3D%20%5Cfrac%7Bn%28No%5C%20girl%29%7D%7Bn%28S%29%7D)
![P(No\ girl) = \frac{1}{8}](https://tex.z-dn.net/?f=P%28No%5C%20girl%29%20%3D%20%5Cfrac%7B1%7D%7B8%7D)
![P(No\ girl) = 0.125](https://tex.z-dn.net/?f=P%28No%5C%20girl%29%20%3D%200.125)