The question reads as follows: "The bombing of London during World War II was studied by statisticians as a Poisson random variable. One of the goals was to determine whether the Germans were bombing randomly or could target specific areas. London was divided into a grid consisting of 576 squares, each of area 0.25 square kilometers, and the number of bombs that landed in each grid square was counted. The total number of bombs that fell was 538. The statisticians found that the number of grid squares on which exactly two bombs fell was 93. What is the expected number of grid squares on which exactly two bombs landed if the bombs were dropped at random over the grid?"

So for my attempt at a solution I have lambda = 538/576 and the probability distribution function P{X=k} = (e^(-lambda(t))*(lambda(t)^k)) / k!

Which leads to : P{X=2} = (e^(-0.934t) x (0.934t)^2) / 2!

My problem is that I don't know what value t should have, any help would be greatly appreciated, thank you!

  • 1
    $\begingroup$ On a historical tangent, the British had double agents working for them who misreported the accuracy of the bombs to the Germans. The Germans then adjusted the flights of their bombs, which resulted in taking the brunt of the attack away from densely populated areas. See, for instance, the Wikipedia article on the V-1 flying bomb. $\endgroup$
    – Théophile
    Nov 25, 2015 at 1:08

1 Answer 1


Consider any particular square. Each bomb has probability $1/576$ of landing in that square, so the probability that exactly two of the $538$ bombs land there is ${538 \choose 2} (1/576)^2 (575/576)^{536} \approx 0.1715517836$. Call this $P_2$. Since expected value is additive, the expected number of squares on which exactly two bombs land is $576 P_2 \approx 98.81382735$.

Despite the preamble, this is not a Poisson distribution problem. It used a binomial distribution with $n = 538$ and $p=1/576$. But WWII statisticians didn't have good calculators available, so they would have used the Poisson approximation to that binomial distribution ($\lambda = np = 538/576$, $P_2 \approx e^{-\lambda} \lambda^2/2! = 0.1714140748$, and again $576 P_2 \approx 98.73450708$).


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