The problem said:
In a certain region, blue cranes are twice as common as whooping cranes. Suppose that the number of eggs laid by a blue crane is a Poisson(! = 3) random variable and the number of eggs laid by whooping crane is a Poisson(! = 5) random variable. You find a crane’s nest that contains 4 eggs. What is the probability that it is a whooping crane’s nest?
Therefore I have:
B: blue cranes
W: whooping carnes
Pr(A)=2P(B) B-Poisson (Lambda=3) W-Poisson (Lambda=5)
I believe that,
P( W=4 | B+W=4) = P( (W=4) (INTERCTION) (B+W=4))/P(B+W=4)=P(B=0)/P(B+W=4) = 0.07385
P(B=0)= e^-5
P(B+W=4)=e^(-7) (7^(4)) * 1/4! = 0.091226
But that is not correct because the book said that the correct answer is: 0.342
If someone can help me with this I will be very happy, thank for all the support.
Answer:
Let X1 be blue crane and X2 be whooping crane
$P(X1 = 4) = \dfrac{e^{-3}.3^4}{4!}$
$P(X2 = 4) = \dfrac{e^{-5}.5^4}{4!}$
$P(X1) = \frac{2}{3}$
$P(X2) = \frac{1}{3}$
P(4 Eggs are from X2) $= \dfrac{P(X2).P(X2 = 4)}{P(X2).P(X2 = 4)+P(X1).P(X1 = 4)}$
$= \dfrac{0.333*0.1754673}{0.333*0.1754673+.6667*0.168031356} = 0.343021158$