Respuesta :
Answer:
a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25
For this case the probability mass function would be given by:
[tex] P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...[/tex]
b) [tex] P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)[/tex]
[tex] P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25[/tex]
[tex] P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875[/tex]
[tex] P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406[/tex]
And adding the values we got:
[tex] P(X \leq 3) =0.25+0.1875+0.1406=0.578[/tex]
c) [tex] P(X \geq 5) = 1-P(X<5) = 1- P(X\leq 4)= 1-[P(X=1) +P(X=2) +P(X=3) +P(X=4)][/tex]
And we can find the individual probabilities:
[tex] P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25[/tex]
[tex] P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875[/tex]
[tex] P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406[/tex]
[tex] P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055[/tex]
[tex] P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316[/tex]
Step-by-step explanation:
Previous concepts
The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"
[tex]P(X=x)=(1-p)^{x-1} p[/tex]
Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:
[tex]X\sim Geo (1-p)[/tex]
Part a
Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25
For this case the probability mass function would be given by:
[tex] P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...[/tex]
Part b
We want this probability:
[tex] P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)[/tex]
We find the individual probabilities like this:
[tex] P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25[/tex]
[tex] P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875[/tex]
[tex] P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406[/tex]
And adding the values we got:
[tex] P(X \leq 3) =0.25+0.1875+0.1406=0.578[/tex]
Part c
For this case we want this probability:
[tex] P(X \geq 5)[/tex]
And we can use the complement rule like this:
[tex] P(X \geq 5) = 1-P(X<5) = 1- P(X\leq 4)= 1-[P(X=1) +P(X=2) +P(X=3) +P(X=4)][/tex]
And we can find the individual probabilities:
[tex] P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25[/tex]
[tex] P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875[/tex]
[tex] P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406[/tex]
[tex] P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055[/tex]
[tex] P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316[/tex]
