More probability Questions!! (1 Viewer)

[Bellow]

The probability that n accidents occur at a given intersection during a year is

P(n)= e^(-2.6) x [(2.6)^n]/n!

(i) Find the probability that no accidents occur at the intersection in a given year. Give your answer correct to three dec. places.
(ii) What is the probability that, in a given ten-year period, there are atleast 2 years in which no accidents occur at the intersection? Give your answer correct to three dec. places.
(iii) By considering values of n for which

[P(n+1)]/[P(n)] >(or equal to) 1

determine the most likely number of accidents in a given one-year period.

(I'll be posting more probability Qs up coz i mite need help)

 

[KFunk]

i) The probability that no accidents occur in a year is equal to P₀ which is equal to e^-2.6

ii) Think of this as a binomial probability question [P₀ + (1 - P₀)]¹⁰. P(zero crashes for at least two years) is equal to 1 - (P(zero crashes one year) + P(no years with zero crashes)):

P(at least two) = 1 - ¹⁰C₁(1 - P₀)⁹(P₀) - (1 - P₀)¹⁰
= 0.167

iii) P_n+1/P_n = n!e^-2.6(2.6)ⁿ⁺¹/(n+1)!e^-2.6(2.6)ⁿ

= 2.6/n+1 ≥ 1 which gives us n ≤ 1.6 (∴ n=1)

This tells us that P₂/P₁ ≥ 1 (i.e. that P₂ > P₁).

It also suggests that for n>1, that P_n+1/P_n ≤ 1 , hence we conclude that P₂ > P₃ > ...

Hence the most likely number of accidents to occur is 2 (since P₂ is the greatest value of P_n).

 

[nit]

oooh Poisson distribution...

 

[SeDaTeD]

mmmm, fish