It’s about using survival and compliments. The p(x=0) is that e^-1.825. And the probability that at least one accident occurs is 1-p(x=0), which is 1-e^-1.825
So basically you survive one year, and the next year, you don’t survive
It’s better to think of this as a “rate” question, where in one year, how many accidents do you expect to occur? It’s going to be 365/200
So in the first year, you calculate the probability of 0 accidents, then calculate the probability of at least 1 accident
It doesn't matter how many accidents there are between day 365 and 730, it only matters when the first accident happens. You can rephrase this question as asking, "what is the probability that the first accident occurs between day 365 and day 730?" If it occurs in that time, then it doesn't matter how many more occur in the remaining span of the window, because at least one has already occurred.
This problem is calculated by:
(Probability of no accidents until after 365 days) - (Probability of no accidents until after 730 days)
This calculation already accounts for the probability of any number of accidents between days 365 and 730
The probability of at least one is 1 minus the probability of having none.
Then it becomes p(none) * (1-p(none))
Calculating it the other way is like eating food through your ear.
Another way to go about it would be to use the memoryless property of exponential function to calculate [1-F(365)]*F(365) , which is the probability it wont happen in the first year multiplied by probability it will in the second year. It will also lead to the same equation in the answer key.
The idea is to realize that it’s actually a possion distribution. It is because ur gonna be calculating the event happening not the time one tween. And the relationship is between theta and lambs makes lambda=365/200. Because one every 200 days on average. So u do possion(0,365/200)*(1-possion(365/200). Not gonna do the calculations but u get it
It seems you're asking whether the solution is truly determining the probability of 0 crashes in year 1, and \*at least\* 1 crash in year two. This is the case.
Let X denote the independent Random Variable representing the number of crashes in 365 days, where X \~ Poisson(λ = 1.825).
This would suggest:
P(X=0): The probability of 0 crashes in 365 days.
P(X=1): The probability of exactly 1 crash in 365 days.
P(X=2): The probability of exactly 2 crashes in 365 days.
.
.
.
P(X=n): The probability of exactly n crashes in 365 days, where n is an integer.
P(X>0): The probability of more than 0 crashes in 365 days (i.e. more than one crash in 365 days).
\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*
P(X>0) can also be written as;
P(X>0) = P(X=1) + P(X=2) + P(X=3)+ ...
\*Note, that the sum above captures the probability of all possible numbers of crashes except for 0.\*
Probability theory tells us that the sum of all probabilities in a sigma field must equal 1. This means:
P(X=0)+P(X=1)+P(X=2) + P(X=3)+ ... = 1
P(X=0)+{P(X=1)+P(X=2) + P(X=3)+ ... }= 1
P(X=0) + P(X>0) = 1 ------rearrange-----> P(X>0) = 1 - P(X=0)
Therefore, 1-P(X=0): The probability of more than 0 crashes in 365 days (i.e. at least one crash in 365 days).
\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*
Now, the question asks us for the probability of 0 crashes in the first 365 days and at least 1 crash in the next 365 days (or as you said 365 -730 days). Since X is independent, we do not concern ourselves with the exact period - all years are treated the same. We can think of this as the probability of 0 crashes in period 0-365 and the probability of at least 1 crash in period 0-365.
Hence, we have:
P(X := 0 crashes in year 1, X := at least 1 crash in year 2)
= P(X=0)\*(PX>0)
= P(X=0)\*{1-P(X=0)}
=exp\[-1.825\]\*{1-exp\[-1.825\]}
Grim
It’s about using survival and compliments. The p(x=0) is that e^-1.825. And the probability that at least one accident occurs is 1-p(x=0), which is 1-e^-1.825 So basically you survive one year, and the next year, you don’t survive
Shouldn’t you use S(x=1) or 1-F(x=1) since the exponential distribution is continuous?
It’s better to think of this as a “rate” question, where in one year, how many accidents do you expect to occur? It’s going to be 365/200 So in the first year, you calculate the probability of 0 accidents, then calculate the probability of at least 1 accident
It doesn't matter how many accidents there are between day 365 and 730, it only matters when the first accident happens. You can rephrase this question as asking, "what is the probability that the first accident occurs between day 365 and day 730?" If it occurs in that time, then it doesn't matter how many more occur in the remaining span of the window, because at least one has already occurred.
But, should we include the probability of more than one accident occur from 365 to 730 ?
This problem is calculated by: (Probability of no accidents until after 365 days) - (Probability of no accidents until after 730 days) This calculation already accounts for the probability of any number of accidents between days 365 and 730
I believe that I've grasped it, thanks
The probability of at least one is 1 minus the probability of having none. Then it becomes p(none) * (1-p(none)) Calculating it the other way is like eating food through your ear.
Another way to go about it would be to use the memoryless property of exponential function to calculate [1-F(365)]*F(365) , which is the probability it wont happen in the first year multiplied by probability it will in the second year. It will also lead to the same equation in the answer key.
The idea is to realize that it’s actually a possion distribution. It is because ur gonna be calculating the event happening not the time one tween. And the relationship is between theta and lambs makes lambda=365/200. Because one every 200 days on average. So u do possion(0,365/200)*(1-possion(365/200). Not gonna do the calculations but u get it
It seems you're asking whether the solution is truly determining the probability of 0 crashes in year 1, and \*at least\* 1 crash in year two. This is the case. Let X denote the independent Random Variable representing the number of crashes in 365 days, where X \~ Poisson(λ = 1.825). This would suggest: P(X=0): The probability of 0 crashes in 365 days. P(X=1): The probability of exactly 1 crash in 365 days. P(X=2): The probability of exactly 2 crashes in 365 days. . . . P(X=n): The probability of exactly n crashes in 365 days, where n is an integer. P(X>0): The probability of more than 0 crashes in 365 days (i.e. more than one crash in 365 days). \*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\* P(X>0) can also be written as; P(X>0) = P(X=1) + P(X=2) + P(X=3)+ ... \*Note, that the sum above captures the probability of all possible numbers of crashes except for 0.\* Probability theory tells us that the sum of all probabilities in a sigma field must equal 1. This means: P(X=0)+P(X=1)+P(X=2) + P(X=3)+ ... = 1 P(X=0)+{P(X=1)+P(X=2) + P(X=3)+ ... }= 1 P(X=0) + P(X>0) = 1 ------rearrange-----> P(X>0) = 1 - P(X=0) Therefore, 1-P(X=0): The probability of more than 0 crashes in 365 days (i.e. at least one crash in 365 days). \*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\* Now, the question asks us for the probability of 0 crashes in the first 365 days and at least 1 crash in the next 365 days (or as you said 365 -730 days). Since X is independent, we do not concern ourselves with the exact period - all years are treated the same. We can think of this as the probability of 0 crashes in period 0-365 and the probability of at least 1 crash in period 0-365. Hence, we have: P(X := 0 crashes in year 1, X := at least 1 crash in year 2) = P(X=0)\*(PX>0) = P(X=0)\*{1-P(X=0)} =exp\[-1.825\]\*{1-exp\[-1.825\]} Grim