Answer:
(a) 4 sample points
(b) See attachment for tree diagram
(c) The probability that no tail is appeared is 1/4
(d) The probability that exactly 1 tail is appeared is 1/2
(e) The probability that 2 tails are appeared is 1/4
(f) The probability that at least 1 tail appeared is 3/4
Step-by-step explanation:
Given
![Coins = 2](https://tex.z-dn.net/?f=Coins%20%3D%202)
Solving (a): Counting principle to determine the number of sample points
We have:
![Coin\ 1 = \{H,T\}](https://tex.z-dn.net/?f=Coin%5C%201%20%3D%20%5C%7BH%2CT%5C%7D)
![Coin\ 2 = \{H,T\}](https://tex.z-dn.net/?f=Coin%5C%202%20%3D%20%5C%7BH%2CT%5C%7D)
To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:
![S = \{HH,HT,TH,TT\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7BHH%2CHT%2CTH%2CTT%5C%7D)
The number of sample points is:
![n(S) = 4](https://tex.z-dn.net/?f=n%28S%29%20%3D%204)
Solving (b): The tree diagram
See attachment for tree diagram
From the tree diagram, the sample space is:
![S = \{HH,HT,TH,TT\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7BHH%2CHT%2CTH%2CTT%5C%7D)
Solving (c): Probability that no tail is appeared
This implies that:
![P(T = 0)](https://tex.z-dn.net/?f=P%28T%20%3D%200%29)
From the sample points, we have:
--- i.e. 1 occurrence where no tail is appeared
So, the probability is:
![P(T = 0) = \frac{n(T = 0)}{n(S)}](https://tex.z-dn.net/?f=P%28T%20%3D%200%29%20%3D%20%5Cfrac%7Bn%28T%20%3D%200%29%7D%7Bn%28S%29%7D)
This gives:
![P(T = 0) = \frac{1}{4}](https://tex.z-dn.net/?f=P%28T%20%3D%200%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
Solving (d): Probability that exactly 1 tail is appeared
This implies that:
![P(T = 1)](https://tex.z-dn.net/?f=P%28T%20%3D%201%29)
From the sample points, we have:
--- i.e. 2 occurrences where exactly 1 tail appeared
So, the probability is:
![P(T = 1) = \frac{n(T = 1)}{n(S)}](https://tex.z-dn.net/?f=P%28T%20%3D%201%29%20%3D%20%5Cfrac%7Bn%28T%20%3D%201%29%7D%7Bn%28S%29%7D)
This gives:
![P(T = 1) = \frac{2}{4}](https://tex.z-dn.net/?f=P%28T%20%3D%201%29%20%3D%20%5Cfrac%7B2%7D%7B4%7D)
![P(T = 1) = \frac{1}{2}](https://tex.z-dn.net/?f=P%28T%20%3D%201%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
Solving (e): Probability that 2 tails appeared
This implies that:
![P(T = 2)](https://tex.z-dn.net/?f=P%28T%20%3D%202%29)
From the sample points, we have:
--- i.e. 1 occurrences where 2 tails appeared
So, the probability is:
![P(T = 2) = \frac{n(T = 2)}{n(S)}](https://tex.z-dn.net/?f=P%28T%20%3D%202%29%20%3D%20%5Cfrac%7Bn%28T%20%3D%202%29%7D%7Bn%28S%29%7D)
This gives:
![P(T = 2) = \frac{1}{4}](https://tex.z-dn.net/?f=P%28T%20%3D%202%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
Solving (f): Probability that at least 1 tail appeared
This implies that:
![P(T \ge 1)](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29)
In (c), we have:
![P(T = 0) = \frac{1}{4}](https://tex.z-dn.net/?f=P%28T%20%3D%200%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
Using the complement rule, we have:
![P(T \ge 1) + P(T = 0) = 1](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29%20%2B%20P%28T%20%3D%200%29%20%3D%201)
Rewrite as:
![P(T \ge 1) = 1-P(T = 0)](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29%20%20%3D%201-P%28T%20%3D%200%29)
Substitute known value
![P(T \ge 1) = 1-\frac{1}{4}](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29%20%20%3D%201-%5Cfrac%7B1%7D%7B4%7D)
Take LCM
![P(T \ge 1) = \frac{4-1}{4}](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29%20%20%3D%20%5Cfrac%7B4-1%7D%7B4%7D)
![P(T \ge 1) = \frac{3}{4}](https://tex.z-dn.net/?f=P%28T%20%5Cge%201%29%20%20%3D%20%5Cfrac%7B3%7D%7B4%7D)