Answer to Question #178370 in Statistics and Probability for anithra

Question

There are two AI (artificial intelligence) programs designed for a certain purpose, Kato and Goly. Both of them are run continuously and simultaneously to check for errors. It was found that the number of errors occur in an hour for Kato follows the Poisson Distribution with the mean of 4, whereas the number of errors occur in an hour for Goly follows the Poisson Distribution with the mean of 7.

Find the probability of the following, give your answer to at least 4 significant figures: a) exactly 3 errors occur in Kato within an hour. (3 marks)

b) at least 2 errors occur in Goly within an hour. (4 marks)

c) less than 6 errors occur in Kato within 3 hours. (5 marks)

d) more than 11 errors occur in Goly within 2.5 hours. (8 marks)

Answer 1

Solution:

Let K denotes the random variable of the number of errors for Kato A.I.

Then, $K\sim Poisson(\lambda)$

with $\lambda=4\ per\ hour$

Also, let G denotes the random variable of the number of errors for Goly A.I.

Then, $G\sim Poisson(\lambda)$

with $\lambda=7\ per\ hour$

(a): $P(K=3)=\dfrac{e^{-4}4^3}{3!}=0.1953$

(b): $P(G\ge2)=1-P(G<2)=1-P(G=0)-P(G=1)$

$=1-\dfrac{e^{-7}7^0}{0!}-\dfrac{e^{-7}7^1}{1!}=0.9927$

(c): For Kato, for three hours, $\lambda=4(3)=12$

$P(K<6)=P(K=0)+P(K=1)+P(K=2)+P(K=3)\\+P(K=4)+P(K=5)$

$=\dfrac{e^{-12}{12}^0}{0!}+\dfrac{e^{-12}{12}^1}{1!}+\dfrac{e^{-12}{12}^2}{2!}+\dfrac{e^{-12}{12}^3}{3!}+\dfrac{e^{-12}{12}^4}{4!}+\dfrac{e^{-12}{12}^5}{5!}$

$=$ 0.02034

(d): For Goly, for 2.5 hours, $\lambda=7(2.5)=17.5$

$P(G>11)=1-P(G\le11) \\=1-[P(G=0)+P(G=1)+P(G=2)+...+P(G=11)] \\=1-[\dfrac{e^{-17.5}{17.5}^0}{0!}+\dfrac{e^{-17.5}{17.5}^1}{1!}+\dfrac{e^{-17.5}{17.5}^2}{2!}+...+\dfrac{e^{-17.5}{17.5}^{11}}{11!}]$

$=$ 0.93160