Answer:
a)
b)
![P(X=0) = (2C0) (0.5)^0 (1-0.5)^(2-0)= 0.25](https://img.qammunity.org/2021/formulas/mathematics/high-school/d2fm3wjn4nta7bqudhfw3fsvjn2y6lhkaw.png)
![P(X=1) = (2C1) (0.5)^1 (1-0.5)^(2-1)= 0.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/er91ra6jj7uox8ax5brn7x9915guuhmfjm.png)
![P(X=2) = (2C2) (0.5)^2 (1-0.5)^(2-2)= 0.25](https://img.qammunity.org/2021/formulas/mathematics/high-school/rucb7pg2i0c6i68v80sh0pwguld42hsork.png)
And we have the following table:
X | 0 | 1 | 2
P(X) | 0.25 | 0.5 | 0.25
Explanation:
Let's define first some notation
H= represent a head for the coin tossed
T= represent tails for the coin tossed
We are going to toss a coin 2 times so then the size of the sample size is
![2^2 = 4](https://img.qammunity.org/2021/formulas/mathematics/high-school/j8ofhbkt7l97f7xyocg088xupkwu98tylt.png)
a. What is the sample space for this chance experiment?
The sampling space on this case is given by:
b. For this chance experiment, give the probability distribution for the random variable of the total number of heads observed.
The possible values for the number of heads are X=0,1,2. If we assume a fair coin then the probability of obtain heads is the same probability of obtain tails and we can find the distribution like this:
![P(X=0) = (2C0) (0.5)^0 (1-0.5)^(2-0)= 0.25](https://img.qammunity.org/2021/formulas/mathematics/high-school/d2fm3wjn4nta7bqudhfw3fsvjn2y6lhkaw.png)
![P(X=1) = (2C1) (0.5)^1 (1-0.5)^(2-1)= 0.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/er91ra6jj7uox8ax5brn7x9915guuhmfjm.png)
![P(X=2) = (2C2) (0.5)^2 (1-0.5)^(2-2)= 0.25](https://img.qammunity.org/2021/formulas/mathematics/high-school/rucb7pg2i0c6i68v80sh0pwguld42hsork.png)
And we have the following table:
X | 0 | 1 | 2
P(X) | 0.25 | 0.5 | 0.25