# Joint and conditional probability with Poisson and Binomial distributions

I'm having a hard time trying to figuring out how to resolve a problem that involves a Poisson distribution and a Binomial distribution.

Let's suppose that the total number of offspring (sons and daughters) in a randomly chosen family can be described as a random variable with a Poisson distribution parameter λ = 1.8. Let's also assume that the probability of having a male child is 0.51, independently in different births. If it is known that the number of daughters in a family is exactly 2, what is the probability that they also have at least one son?

I defined the following random variables

$$X$$: Total number of offspring, then $$X \sim Poisson(\lambda = 1.8)$$

$$Y$$: Total number of male child

$$Y | X = x \sim Binomial(n = 2, p = 0.51)$$

I intuitively could say that the problem is asking us for

$$P(X \ge 3, Y \ge 1) = P(Y \ge 1 | X \ge 3) P(X \ge 3)$$

At this point I'm stuck, and I don't even know if I'm getting right the probability that is asked.

• Do you mean $$Y | X = x \sim Binomial(n = x, p = 0.51)$$ ? In your formulation, I think the solution is to find $P(Y\ge1 | X-Y = 2)$ . Your approach does not condition on having exactly two daughters. Commented Nov 15, 2023 at 22:53
• There is a simple solution to this puzzle. Think about this: you are in a world every family has at leats two children and every other child which may or may not exist is a son. Commented Nov 15, 2023 at 23:01

This problem has its roots on compound Poisson process. The correct way of approaching it is to express the number of boys $$X$$ in a household as \begin{align*} X = D_1 + \cdots + D_N, \end{align*} where $$N \sim \text{Poisson}(1.8)$$, $$D_1, D_2, \ldots \text{ i.i.d. } \sim B(1, 0.51)$$, which are also independent of $$N$$ (you didn't explicitly state this condition, but I infer it should hold to solve this problem). With these notations, the probability of your interest is

\begin{align*} & P[X \geq 1 | N - X = 2] = \frac{P[X \geq 1, N - X = 2]}{P[N - X = 2]} \\ =& \frac{P[N - X = 2] - P[X = 0, N - X = 2]}{P[N - X = 2]} \\ =& 1 - \frac{P[X = 0, N = 2]}{P[N - X = 2]}, \tag{1}\label{1} \end{align*} where \begin{align*} P[X = 0, N = 2] = P[D_1 + D_2 = 0, N = 2] = 0.49^2 \times e^{-1.8} \times \frac{1.8^2}{2!} = 0.06429. \tag{2}\label{2} \end{align*}

To determine $$P[N - X = 2]$$, apply the law of total probability as follows: \begin{align*} & P[N - X = 2] = \sum_{n = 0}^\infty P[N - X = 2, N = n] \\ =& \sum_{n = 0}^\infty P[X = n - 2, N = n] \\ =& \sum_{n = 2}^\infty P[X = n - 2, N = n] \\ =& \sum_{n = 2}^\infty P[X = n - 2]P[N = n] \\ =& \sum_{n = 2}^\infty \binom{n}{n - 2}p^{n - 2}(1 - p)^2\lambda^n e^{-\lambda}/n! \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-\lambda}\sum_{n = 2}^\infty \frac{(p\lambda)^{n - 2}}{(n - 2)!} \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-\lambda}\sum_{k = 0}^\infty \frac{(p\lambda)^k}{k!} \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-(1 - p)\lambda} = \frac{1}{2} \times 1.8^2 \times 0.49^2 \times e^{-0.49 \times 1.8} = 0.1610. \tag{3}\label{3} \end{align*}

Substituting $$\eqref{2}$$ and $$\eqref{3}$$ into $$\eqref{1}$$ gives \begin{align*} P[X \geq 1 | N - X = 2] = 1 - \frac{0.06429}{0.1610} = 0.6007. \end{align*}

• thanks for your kind help, now I understand more about conditional probability. I'm still don't totally get this $$\frac{ P \left[ X \ge 1, N - X = 2 \right] }{P \left[ N - X = 2 \right] } = \frac{P \left[ N- X = 2 \right] - P \left[ X = 0, N - X = 2 \right] }{P \left[ N - X = 2 \right] }$$ because, I was doing it on paper and I've got $$\frac{ P \left[ X \ge 1, N - X = 2 \right] }{ P \left[ N - X = 2 \right] } = \frac{1 - P \left[ X = 0, N - X = 2 \right] }{P \left[ N - X = 2 \right] }$$ Commented Nov 19, 2023 at 12:34
• @CarloRock I believe you probably know $P(A) = P(A \cap B) + P(A \cap B^c) \neq 1$. Now to understand the numerator of the expression you wrote down, just take $A = \{N - X = 2\}$ and $B = \{X \geq 1\}$. Commented Nov 19, 2023 at 17:16
• Now I totally understand the whole process, thanks! Commented Dec 9, 2023 at 21:24

### Simple and short

I started below with some long computation starting from your erroneous $$P(X \ge 3, Y \ge 1)$$. Trying to make an intuitive interpretation of the end result it made me realize that we can view these number of births of boys and girls as two separate independent Poisson distributed variables. So the formula can be written very simple as for independent events we have $$P(A|B) = P(A)$$ $$\text{P(more than k boys | given m girls)} = \text{P(more than k boys)}$$

and solving with the Poisson distribution with rate $$0.51 \cdot \lambda$$ you get

$$\text{P(more than 1 boys)} = 1-e^{- 0.51 \cdot \lambda} = 0.6006831$$

### Longer part

$$P(X \ge 3, Y \ge 1)$$

This is the wrong event. You need to have $$X-Y = 2$$ which is not satisfied for every case in $$X \ge 3, Y \ge 1$$.

Alternatively you could look for the probability

$$P(X \ge 3, X-Y = 2) = \sum_{k=3}^{\infty} P(X=k)\cdot P(Y = k-2) = \sum_{k=3}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} {k\choose 2} p^{k-2}q^2$$

where $$p=0.51$$ and $$q = 0.49$$

By rearranging some terms the sum can be rewritten in terms of a Poisson distribution with parameter $$\lambda p$$

$$\begin{array}{} \sum_{k=3}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} {k\choose 2} p^{k-2}q^2 &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \sum_{k=3}^{\infty} \frac{(\lambda \cdot p)^{k-2} e^{-(\lambda \cdot p)}}{(k-2)!} \\ &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \sum_{k=1}^{\infty} \frac{(\lambda \cdot p)^{k} e^{-(\lambda \cdot p)}}{k!} \\ &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \left( 1- \sum_{k=0}^{0} \frac{(\lambda \cdot p)^k e^{-(\lambda \cdot p)}}{k!} \right) \\ &=& \frac{\lambda^2 q^2}{2} \left( e^{\lambda \cdot (p-1)} - e^{-\lambda}\right)\\ &=& 0.09671746 \end{array}$$

Along with the probability for $$P(X-Y = 2)$$ which is like computing the Poisson distribution with parameter $$q \cdot \lambda$$ giving $$\frac{(\lambda \cdot q)^2 e^{-(\lambda \cdot q)}}{2!} = 0.1610124$$

And

$$P(X \ge 3| X-Y = 2) = \frac{P(X \ge 3, X-Y = 2)}{P(X-Y = 2)} = \frac{0.09671746}{0.1610124}$$

$$P(X = k, X-Y =2) = \frac{\lambda^2 q^2}{2} f(\lambda \cdot p, k-2)$$
where $$f$$ is the probability mass function for the Poisson distribution.
• Here is a justification for the interpretation of the compund poisson process (which is in general not poisson distributed) as a poisson variable with $\lambda = 1.8 * 0.51$ (which is intuitive) math.stackexchange.com/questions/4252625/… Commented Nov 16, 2023 at 22:12