Respuesta :
Answer:
a) 0.164 = 16.4% probability that a disk has exactly one missing pulse
b) 0.017 = 1.7% probability that a disk has at least two missing pulses
c) 0.671 = 67.1% probability that neither contains a missing pulse
Step-by-step explanation:
To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!} [/tex]
In which
x is the number of sucesses
[tex] e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Poisson mean:
[tex]\mu = 0.2[/tex]
a. What is the probability that a disk has exactly one missing pulse?
One disk, so Poisson.
This is P(X = 1).
[tex]P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164 [/tex]
0.164 = 16.4% probability that a disk has exactly one missing pulse
b. What is the probability that a disk has at least two missing pulses?
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!} [/tex]
[tex]P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819[/tex]
[tex]P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164 [/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017[/tex]
0.017 = 1.7% probability that a disk has at least two missing pulses
c. If two disks are independently selected, what is the probability that neither contains a missing pulse?
Two disks, so binomial with n = 2.
A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so [tex]p = 0.181[/tex]
We want to find P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671[/tex]
0.671 = 67.1% probability that neither contains a missing pulse
