Let $X$ denotes the Number of eggs; poisson-distributed with parameter $\lambda$. Each egg hatches independently of the other eggs with probability $p$.
$Y$ denotes the number of the "developed" insects.
What's the probability that $k$ eggs laid given that $n$ insects developed ?
Obviously we have to use the formula of conditional-probability.
So I think that we have to start with:
$$ P( X=k | Y=n ) = \frac{P( {X=k} \cap {Y=n} )}{P ( Y = n )} $$ But I'm not sure. I know that $P$( $X$ = $k$ ) = $e^{-\lambda}$ $\frac{\lambda^k}{k!}$.
Best Answer
So we know that: $$ P( Y=n | X=k ) = {k \choose n} * p^n * (1-p)^{k-n} $$ We also know: $$ P (X = k) = e^{-\lambda} \frac{\lambda^k}{k!} $$ Moreover we know: $$ P(Y=n) = e^{-\lambda*p} \frac{(p*\lambda)^n}{n!} $$ We begin with: $$ P( X=k | Y=n ) = \frac{P( {X=k} \cap {Y=n} )}{P ( Y = n )} = \frac{P( Y=n | X=k )* P(X=k)}{P(Y=n)} $$ Now we use the equalities ( above ) : We get : $$ P( X=k | Y=n ) = e^{-\lambda(1-p)} * \frac{ (\lambda * ( 1 - p ))^{k-n}}{(k-n)!} $$ Am I right or where is my mistake? Is this the probability?