Question

Radio blackouts are among the most common space weather events to affect Earth. Minor radio blackouts occur, on average, twice per year. Use a Poisson process to model the phenomenon. (a) Given that 3 events have occurred in the first half of the year, what is that probability that 3 events will happen in the rest of the year? (b) How many radio blackouts do you expect to see in two years? (c) Starting from now, how long must we wait so that the probability of seeing the next radio blackout is at least 0.5? (d) Find the probability that the time to the 4th blackout is at most 2 years.

Answer 1

Answer:

a.  P(x = 3) = 0.061313

b. The expected number of radio blackouts = 4

c.  It is not possible to determine the next radio blackout

d. P(x = 4) = 0.19537

Step-by-step explanation:

From the given information:

Using a Poisson process to model the phenomenon:

Since minor radio blackouts occur, on average, twice per year.

Then:

$\lambda = 2 \ years$

From question (a)

time t = 1/2 (i.e half of the year)

Let x be the random variable that denotes the probability that 3 events will happen in the rest of the year.

Then:

$P(x = 3) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!}$

$P(x = 3) = \dfrac{e^{-2* 0.5} \times (2 * 0.5)^3}{3!}$

P(x = 3) = 0.061313

b).

The number of radio blackouts  we expect to see in two years can be estimated as follows:

we know that:

t = 2 years

E(x) = λ × t

E(x) = 2 × 2

E(x) = 4

The expected number of radio blackouts = 4

c).

The amount of time we need to wait given that the probability of seeing the next radio blackout is at least 0.5 is as follows:

Here, the time (t) = ???

Thus:

P(x =1) = 0.5

$P(x = 1) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5$

$\dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5$

Thus, it is not possible to determine the probability (50%) of seeing the next radio blackout.

d)

The probability that the time to the 4th blackout is at most 2 years can be computed as follows:

Here;

x =4 , t = 2

Thus:

$P(x = 4) = \dfrac{e^{-4 } \times (4)^4}{4!}$

P(x = 4) = 0.19537