Answer :
Answer:
[tex] f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3[/tex]
1) [tex] f(x_i) \geq 0, \forall x_i[/tex]
2) [tex] sum_{i=1}^n P(X_i) =1[/tex]
We can find the individual probabilities and we got:
[tex] f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923[/tex]
[tex] f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307[/tex]
[tex] f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770[/tex]
And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.
[tex] P(X \leq 1) = P(X=1) =0.6923[/tex]
[tex] P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077[/tex]
[tex] P(2 <X<10) = P(X=3) =0.0770[/tex]
Step-by-step explanation:
For this case we have the following density function:
[tex] f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3[/tex]
In order to satisfty that this function is a probability mass function we need to check two conditions:
1) [tex] f(x_i) \geq 0, \forall x_i[/tex]
2) [tex] sum_{i=1}^n P(X_i) =1[/tex]
We can find the individual probabilities and we got:
[tex] f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923[/tex]
[tex] f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307[/tex]
[tex] f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770[/tex]
And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.
And if we want to find the following probabilities:
[tex] P(X \leq 1) = P(X=1) =0.6923[/tex]
[tex] P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077[/tex]
[tex] P(2 <X<10) = P(X=3) =0.0770[/tex]