How can i calculate the following statistics inside a round area of meteors fall(tricky question)? (details inside)

A rain of meteors is falling on a round area, with a radius of 10 km, so that each meteor is falling in a random please inside the are, without any dependancy on other meteor falls. assuming that the number of meteors that fall inside the round area is a poisson random variable with the parameter 2, and that the three assumptions of the poisson prccess are valid in the area where the meteors fall:

1)what are the odds thatexactly 3 meteors will fall inside that area?

2)inside the round area we mark a round circle with the radius of 6 km, inside the round area mentioned above. what are the odds that inside the little circle mentioned(the 6km), won't fall any meteor?

3)if we know that in the biggercircle fell exactly 4 meteors(the 10 km), what are the odds that exactly one of them fell inside the small area(the 6km) inside the 10 km?

just to clarify, the small circle(6km radius) is inside the bigger circle(10km radius)

1 Answer
MattyMatty
May 6, 2018

#1) 0.180447#
#2) 0.48675#
#3) 0.37749#

Explanation:

#"Poisson : the odds for k events in a time span t is"#
#((lambda * t)^k exp(-lambda*t)) / (k!)#
#"Here we have no further specification of the time span, so we"#
#"take t=1, "lambda = 2.#
#=> P["k events"] = (2^k * exp(-2)) / (k!)#

#"1) "P["3 events"] = (2^3 * exp(-2)) / (3!) = (4/3) e^-2 = 0.180447#

#"2) "(6/10)^2 = 36/100 = 0.36 " is the fraction surface of the"#
#"smaller circle compared to the bigger one."#
#"The odds that a in the bigger circle (BC) falling meteor falls in"#
#"the smaller circle (SC) are 0.36 as such."#
#=> P["0 events in SC"] = P["0 events in BC"] + 0.64 * P["1 event in BC"] + 0.64^2 * P["2 events in BC"] + ...#
#= sum_{i=0}^oo P["i events in BC"] * 0.64^i#
#= sum_{i=0}^oo ((2^i * exp(-2))/(i!)) * 0.64^i#
#= exp(-2) sum_{i=0}^oo (1.28^i / (i!))#
#= exp(-2) exp(1.28)#
#= exp(1.28 - 2)#
#= exp(-0.72)#
#= 0.48675#

#"3) P[1 meteor in SC | 4 meteors in BC] ?"#
#"We have to apply the binomial distribution with"#
#"n=4 ; p=0.36 ; k=1"#
#= C(4,1) * 0.36 * 0.64^3#
#( C(n,k) = (n!)/((n-k)! k!) = " combinations ")#
#= 4 * 0.36 * 0.64^3#
#= 1.44 * 0.64^3#
#= 0.37749#

Related questions
Impact of this question
1241 views around the world
You can reuse this answer
Creative Commons License