There are $n$ eggs, each of which hatches a chick with probability $p$ (independently). Each of these chicks survives with probability $r$, independently. What is the distribution of the number of chicks that hatch? What is the distribution of the number of chicks that survive? (Give the PMFs; also give the names of the distributions and their parameters, if applicable.)

I'm trying to answer this as mathematically rigorously as possible.

Let $X$ be the number of chicks that hatch and let $Y$ be the number of chicks that survive. It's obvious that $X \sim Bin(n, p)$, so the PMF of $X$ is: $$p_X(x) = P(X=x) = {n\choose x}p^x(1-p)^{n-x}$$ I tried to define and calculate the conditional distribution of $X\,|\,Y$, where the number of chicks that survive is conditional on the number of chicks that hatch. $$p_{Y|X}(y|x) = P(Y=y|X=x) = \frac{P(Y=y, X=x)}{P(X=x)}$$.

I couldn't figure out how to express $P(Y=y,X=x)$. $P(Y=y)P(X=x|Y=y)$ didn't seem to make much sense.


1 Answer 1


$X$ is $\text{Bin}(n,p)$, not $r$. And, given number of chicks ($X=x$), the number of surviving chicks is already a Bernoulli with $n=x$ and $p=r$. You don't need to use conditional probability formula for that. But, the question is to find "distribution of number of chicks that survive", i.e. $P_Y(y)$, instead of $P_{Y|X}(y)$.

An easy way to think is each egg has $pr$ probability of becoming a surviving chicken, let's call it as $Y_i$, and $Y=\sum Y_i$, therefore again Binomial distributed, i.e. $\text{Bin}(n, pr)$.

  • $\begingroup$ Thanks @gunes! I did solve the problem the way that you suggested beforehand. I guess I mainly asked the question to see if there's a way to define the problem conditionally as a mental exercise. $\endgroup$
    – Brian Ko
    Nov 2, 2020 at 12:52
  • $\begingroup$ Well, since the data generation process goes from $X$ to $Y$, the easiest thing to come up is $p(y|x)$. You can write $p(y,x)$ but the straightforward thinking leads from $p(y|x)p(x)$. $\endgroup$
    – gunes
    Nov 2, 2020 at 13:10

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