Respuesta :
Answer:
a) [tex] P(X \geq 1) = 1-P(X<1) = 1-P(X=0)[/tex]
[tex]P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}[/tex]
And replacing we got:
[tex] P(X \geq 1) =1 -1x10^{-18} \approx 1[/tex]
b) [tex]P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355[/tex]
[tex]P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047[/tex]
[tex]P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517[/tex]
And adding we got:
[tex] P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992[/tex]
c) [tex] P(X \leq 8) =1 -P(X>8) = 1-P(X=9)[/tex]
[tex]P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517[/tex]
And replacing we got:
[tex] P(X \leq 8)= 1-0.913517=0.086483[/tex]
Step-by-step explanation:
Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:
[tex]X \sim Binom(n=9, p=0.99)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
We want to find this probability:
[tex] P(X \geq 1) = 1-P(X<1) = 1-P(X=0)[/tex]
[tex]P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}[/tex]
And replacing we got:
[tex] P(X \geq 1) =1 -1x10^{-18} \approx 1[/tex]
Part b
[tex] P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9) [/tex]
[tex]P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355[/tex]
[tex]P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047[/tex]
[tex]P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517[/tex]
And adding we got:
[tex] P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992[/tex]
Part c
[tex] P(X \leq 8) =1 -P(X>8) = 1-P(X=9)[/tex]
[tex]P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517[/tex]
And replacing we got:
[tex] P(X \leq 8)= 1-0.913517=0.086483[/tex]
