Birds and laying eggs probability calculator question

Question

The number of eggs, $Y$, laid by female birds of this species during the nesting season is modelled by a Poisson distribution with mean $\lambda$. You are given that $\mathbb P(Y \geq 2) = 0.80085$, correct to 5 decimal places.

i) Determine the value of $\lambda$

Our answer:

Given: $\mathbb P(Y < 2) ~=~ 1-0.80085 ~=~ 0.19915$

Thus: $\mathbb P(Y=0) + \mathbb P(Y=1) = 0.19915$

Then: $e^{-m} +me^{-m} =0.19915$

Therefore: $\lambda = 3$

ii) Calculate the probability that two randomly chosen birds lay a total of two eggs between them.

The difference ways of this happening I wrote was for the two birds

$\mathsf P(2,0), \mathsf P(0,2) , \mathsf P(1,1)$

so, $2\mathbb P(X=2)^2 + \mathbb P(X=1)^2 =0.123$

Answer is actually $0.0446$.

iii) Given that the two birds lay a total of two eggs between them, calculate the probability that they each lay one egg.

Don't understand :(

Sorry if the formatting and presentation is unclear. Be sure to ask if you need clarifications

Answer 1

Your answer to i) is correct.

Your answer to ii) should be $2\mathbb{P}[Y=2]*\mathbb{P}[Y=0]+\mathbb{P}[Y=1]^{2}$. That is, the two events that the number of eggs is $(2,0)$ and the event that the number of eggs is $(1,1)$.

About iii), the question asks for a probability of $(1,1)$ given that either $(1,1)$, or $(2,0)$ or $(0,2)$ occurred. That is, the answer is $\mathbb{P}[Y=1]^{2}/(2\mathbb{P}[Y=2]*\mathbb{P}[Y=0]+\mathbb{P}[Y=1]^{2})$.