Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let X and Y be the accident counts at the two intersections and condition on their total S = X + Y = 10.

X∼Poisson⁡(6),Y∼Poisson⁡(6),X⊥YX\sim\operatorname{Poisson}(6),\quad Y\sim\operatorname{Poisson}(6),\quad X\perp Y
S=X+Y∼Poisson⁡(12)S=X+Y\sim\operatorname{Poisson}(12)

Model

Model

For x = 0, 1, …, 10 the event {X = x, S = 10} is the event {X = x, Y = 10 − x}. Dividing by P(S = 10) cancels every exponential factor and leaves a binomial distribution whose success probability is the share of the total mean contributed by X.

Pr⁡(X=x∣S=10)=e−66xx!⋅e−6610−x(10−x)!e−12121010!=(10x)(12)10\Pr(X=x\mid S=10)=\frac{\frac{e^{-6}6^{x}}{x!}\cdot\frac{e^{-6}6^{10-x}}{(10-x)!}}{\frac{e^{-12}12^{10}}{10!}}=\binom{10}{x}\left(\tfrac{1}{2}\right)^{10}

Compute

Compute

Add the binomial probabilities for 8, 9, and 10 accidents at X.

Pr⁡(X≥8∣S=10)=(108)+(109)+(1010)210=45+10+11024=561024=0.0546875\Pr(X\ge 8\mid S=10)=\frac{\binom{10}{8}+\binom{10}{9}+\binom{10}{10}}{2^{10}}=\frac{45+10+1}{1024}=\frac{56}{1024}=0.0546875

Answer

Answer

The conditional probability rounds to 0.055.

0.055(B)\boxed{0.055\quad\text{(B)}}