I'm trying to solve the following problem:
The number of fruits in a pudding follows a Poisson distribution with parameter $\lambda =60$. A person grabs all fruits from the pudding and shares them with a friend with the following procedure.
He tosses a balanced coin and gives a fruit to his friend if the result is heads, and gives the fruit to himself if the result is tails.
i) Find the distribution of the number of fruits the friend gets.
ii) Find the probability that each of them gets $30$ fruits.
iii) Find the probability that the pudding has more than $60$ fruits, given that the friend got $30$ fruits.
My attempt:
If $X$ is the random variable counting the number of fruits in the pudding, we have $X\sim \operatorname{Po}(60)$. If $Y$ is the random variable counting the number of fruits the friend gets, since the event of getting a fruit has probability $\frac{1}{2}$ (because the coin is balanced) we get $Y\sim \operatorname{Po}(30)$. This answers i).
If $Z$ is the random variable counting the number of fruits he gets, with the same reasoning we find $Z\sim \operatorname{Po}(30)$. Now, I think that $Y$ and $Z$ are independent (I'm not really sure), so $\mathbb{P}(Y=30,Z=30)=\mathbb{P}(Y=30)\cdot \mathbb{P}(Z=30)=\frac{30^{30}}{30!}e^{-30}\cdot \frac{30^{30}}{30!}e^{-30}$. So this should answer ii).
I have no idea about iii).
Is this correct? How can I complete the proof?