Respuesta :
Answer:
a) [tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
b) [tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]
c) [tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]
Step-by-step explanation:
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 os trials until the first success, we know that X follows this distribution:
[tex]X\sim Geo (1-p)[/tex]
Part a
For this case we want this probability
[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
Part b
For this case we want this probability:
[tex]P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)[/tex]
If we find the individual probabilities we got:
[tex]P(X=1)=(1-0.6)^{1-1} 0.6 = 0.6[/tex]
[tex]P(X=2)=(1-0.6)^{2-1} 0.6 = 0.24[/tex]
[tex]P(X=3)=(1-0.6)^{3-1} 0.6 = 0.096[/tex]
[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
And replacing we have:
[tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]
Part c
For this case at least 4 trials means that the random variable X needs to be 4 or more
[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[P(X=1)+P(X=2)+P(X=3)][/tex]
And we found already the probabilities P(X=1),P(X=2) and P(X=3) so we just need to replace:
[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]
