Answer:
a. Â P(x = 3) = 0.061313
b. The expected number of radio blackouts = 4
c. Â 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:
[tex]\lambda = 2 \ years[/tex]
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:
[tex]P(x = 3) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!}[/tex]
[tex]P(x = 3) = \dfrac{e^{-2* 0.5} \times (2 * 0.5)^3}{3!}[/tex]
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
[tex]P(x = 1) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5[/tex]
[tex]\dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5[/tex]
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:
[tex]P(x = 4) = \dfrac{e^{-4 } \times (4)^4}{4!}[/tex]
P(x = 4) = 0.19537
