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Synthemesc90
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Homework Statement
If people entering the engineering building following a PP (9/hora) and you know that between 10:00 am and 11:00 am came exactly 100 persons, what is the probability that between 10:00 am and 10:20 am entered less than 20 people
If people entering the engineering building following a PP (9/hora) and you know that between 10:00 am and 11:00 am came exactly 80 persons, what is the probability that between 10:00 am and 10:20 am entered less than 10 people?
If people entering the engineering building following a PP (9/hora) and you know that between 10:00 am and 12:00 am came exactly 2 persons, what is the probability that between 8:00 am and 11:00 am 1 people have entered?
Homework Equations
The Attempt at a Solution
solution:
P {N (1 / 3) = 20 / N (1) = 100} = (P {N (1 / 3) = 20, N (1) = 100}) / (P {N (1) = 100} )
= (P {N (1 / 3) = 20} P {N (1)-N (1 / 3 )})/( P {N (1) = 100})
P {N (1) = 100} = (9 ^ 100 e ^ (-9)) / 100! = 3.51 * 10-67
P {N (1 / 3) = 20} = (3 ^ 20 e ^ (-3)) / 20! = 7.14 * 10-11
P {N (2 / 3) = 80} = (6 ^ 80 e ^ (-6)) / 80! = 6.19 * 10-60
(7.14 * 〖10〗 ^ (-11) * 6.19 * 〖10〗 ^ (-60)) / (3.51 * 〖10〗 ^ (-67)) = 1.26 * 10-3
* For n ≤ 20
Σ_ (i = 0) ^ 20 ▒ 〖((3 ^ i / i! 6 ^ (100-i) / (100-i 9 ^ )!))/( 100/100!) = 2.37 * 〖10〗 ^ (-3)〗
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P {N (1 / 3) = 10 / N (1) = 80} = (P {N (1 / 3) = 10, N (1) = 80}) / (P {N (1) = 80} )
= (P {N (1 / 3) = 10} P {N (1)-N (1 / 3 )})/( P {N (1) = 80})
P {N (1) = 80} = (9 ^ 80 e ^ (-9)) / 80! = 3.77 * 10-47
P {N (1 / 3) = 10} = (3 ^ 10 e ^ (-3)) / 10! = 8.1 * 10-4
P {N (2 / 3) = 70} = (6 ^ 70 e ^ (-6)) / 70! = 6.12 * 10-49
(8.1 * 〖10〗 ^ (-4) * 6.12 * 〖10〗 ^ (-49)) / (3.77 * 〖10〗 ^ (-47)) = 1.32 * 10-5
* For n ≤ 10
Σ_ (i = 0) ^ 10 ▒ 〖((3 ^ i / i! 6 ^ (80-i) / (80-i )!))/( 9 ^ 80/80!) = 1.80 * 〖10〗 ^ (-5)〗
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For the third point, I don't know how take the interval, any advice?