Respuesta :
Answer:
0.0509 is the probability that the article will have no mistakes.
Step-by-step explanation:
We are given the following in the question:
P(Typist A) = P(Typist B) = 0.5
P(typist A makes an error) = 2.4
The number of errors made by typist A is a Poisson distribution.
Formula:
[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]
[tex]\lambda_1 = 2.4[/tex]
We have to evaluate probability for no mistakes.
P(Typist A makes no mistake) =
[tex]P(x=0) = \dfrac{(2.4)^0e^{-2.4}}{0!} =0.0907[/tex]
P(typist B makes an error) = 4.5
The number of errors made by typist B is a Poisson distribution.
[tex]\lambda_2 = 4.5[/tex]
We have to evaluate probability for no mistakes.
P(Typist B makes no mistake) =
[tex]P(x=0) = \dfrac{(4.5)^0e^{-4,.5}}{0!} =0.0111[/tex]
P(article will have no error) =
P(Typist A)P(Typist A makes no mistake) + P(Typist B)P(Typist B makes no mistake)
[tex]=0.5\times 0.0907 + 0.5\times 0.0111\\=0.0509[/tex]
0.0509 is the probability that the article will have no mistakes.
