Hundred archers and one target

[StrangeCoin]

Every archer hits the target every time and they shoot at it for ten seconds with the speed of 1 arrow per 3 seconds exactly, but they all start shooting at different random times within the first three seconds, so their arrows hit the target at different times. What is likely number of arrows to hit the target in any given one second time interval after the first three seconds warmup? Thanks.

 

[FactChecker]

This is a Poisson process with an average rate of 100/3 = 33.333 arrow hits per second. So the expected number of arrow hits in a second is 33.333.

 

[verty]

This is a Poisson process with an average rate of 100/3 = 33.333 arrow hits per second. So the expected number of arrow hits in a second is 33.333.

Is it not a uniform distribution of 100 arrows over 3 seconds?

And by the way, StrangeCoin seems to be asking homework questions in the math section...

 

[StrangeCoin]

This is a Poisson process with an average rate of 100/3 = 33.333 arrow hits per second. So the expected number of arrow hits in a second is 33.333.

I see. I made a mistake. "Warmup time" when each archer starts shooting at random point in time, is supposed to be within first one second or two seconds, not three. Say two seconds, so maybe it's 2 out of 3 = 67 arrows per second would hit the target. Doesn't sound right. It must be 33% in any case, doesn't it?

 

[Hornbein]

Every archer hits the target every time and they shoot at it for ten seconds with the speed of 1 arrow per 3 seconds exactly, but they all start shooting at different random times within the first three seconds, so their arrows hit the target at different times. What is likely number of arrows to hit the target in any given one second time interval after the first three seconds warmup? Thanks.

33.3.

"they all start shooting at different random times " We don't know the distribution but in probability problems if they don't say then it is always uniform and independent. So we get a binomial distribution with n = 100 and p = 1/3. The number of hits during the seven one-second intervals are not independent. H(n) =H(n+3) and H(n)+H(n+1)+H(n+2)=100 so there are only two degrees of freedom. H(2) is a binomial with n=100-H(1) and p = n/2