Respuesta :
Answer:
a) There is a 98.17% probability that a randomly selected page has at least one typo on it.
b) There is a 9.16% probability that a randomly selected page has at most one typo on it.
Step-by-step explanation:
Since we only have the mean, we can solve this problem by a 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
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that [tex]\mu = 4[/tex]
(a) What is the probability that a randomly selected page has at least one typo on it?
Thats is [tex]P(X \geq 1)[/tex]. Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:
[tex]P(X < 1) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X < 1)[/tex]
In which
[tex]P(X < 1) = P(X = 0)[/tex].
So
[tex]P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183[/tex]
[tex]P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817[/tex]
There is a 98.17% probability that a randomly selected page has at least one typo on it.
(b) What is the probability that a randomly selected page has at most one typo on it?
This is [tex]P = P(X = 0) + P(X = 1)[/tex]. So:
[tex]P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183[/tex]
[tex]P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733[/tex]
[tex]P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916[/tex]
There is a 9.16% probability that a randomly selected page has at most one typo on it.
