In a particular state, a major ood happens once every 3 years on average. If you

Question

In a particular state, a major ood happens once every 3 years on average. If you can assume oods are independent, and that two oods cannot happen simultaneously,nd the following:
(a) The probability that a ood happens within the rst 5 years.
(b) The probability that a ood happens after the rst 2 years.
(c) If there has been no ood for 4 years, what is theprobability there is a ood in the next 2 years?
(d) The expected time until a ood happens, and the standard deviation.
(e) Find the 50th percentile of the time until a ood.

Explanation / Answer

Solution

Back-up Theory

Occurrence of floods is assumed to follow Exponential Distribution.

If X ~ Exponential with parameter (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/)e-x/, 0 x < ………………………………(1)

CDF (cumulative distribution function), F(t) = P(X 1) = 1- e-x/ …………………(2)

From (2), P(X > t) = e-x/ ……………………………………………………………(3)

Now to work out the solution,

In the given problem, = 3 years.

Part (a)

We want probability that a ood happens within the rst 5 years.
= P(X 5) = 1- e-5/3 [vide (2) under Back-up Theory]

= 1 – 0.1889 = 0.8111 ANSWER

Part (b)

The probability that a ood happens after the rst 2 years =>

P(X > 2) = e-2/3 [vide (3) under Back-up Theory]

= 0.5131 ANSWER

Part (c)

If there has been no ood for 4 years, what is the probability there is a ood in the next 2 years?

Let A the event that there has been no ood for 4 years and B be the event that

there is a ood in the next 2 years.

We want P(B/A)
By definition of conditional probability, P(B/A) = P(BA)/P(A).

Now, (BA) => no flood in 4 years and flood in next 2 years. So,

P(BA) = P(no flood in 4 years) x P(flood in next 2 years)

                                         [given floods are independent]

= P(X > 4) x P(X 2)

= e-4/3 x (1 - e-2/3)[vide (3) and (2) under Back-up Theory]

P(A) = e-4/3

Hence, P(B/A) = { e-4/3 x (1 - e-2/3)}/(e-4/3) = (1 - e-2/3) = 1 – 0.5131 = 0.4849 ANSWER

Part (d)

The expected time until a ood happens.

Let t = time until a ood happens. Then

t = 1, if there is no flood in 1 year

t = 2, if there is no flood in 2 years

t = 3, if there is no flood in 3 years

Mathematically, t may go to infinity.

So, expected time until a ood happens

= (1. e-1/3) + (2. e-2/3) + (3. e-3/3) + …… to infinity = (e-1/3)/{1 – (e-1/3)}2 = 0.7165/0.0804 = 8.9 ~ 9 years ANSWER

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