Probability that all the children are girls given a condition

[  maths  probability  ] Last Updated: 07 Jun 2021

Question

A family has \(n\) children, \(n \geq 2\). What is the probability that all of their children are girls for the following cases:

a) We ask the father: “Do you have at least one daughter?” He responds “Yes!”

b) We pick one of them at random and find out that she is a girl

c) We ask the father, “Do you have at least one daughter named Lilia?” He replies, “Yes!”

Here you can assume that if a child is a girl, her name will be Lilia with probability \(\alpha \ll 1\) independently from other children’s names. If the child is a boy, his name will not be Lilia.

Answer

Let us chart down what we know:

\(E_1 \Rightarrow \text{Event that atleast one of the children is daughter}\) \(E_2 \Rightarrow \text{Event that the random child we picked was a girl}\) \(E_3 \Rightarrow \text{Event that atleast one of the children is a daughter named Lilia}\) \(G_n \Rightarrow \text{Event that all the } n \text{ children are girls}\)

To find,

\[P(G_n|E_1) \Rightarrow \text{Probability that all the children are girls given } E_1\] \[P(G_n|E_2) \Rightarrow \text{Probability that all the children are girls given } E_2\]

Let,

\[G_i \Rightarrow \text{Event that all the } i \text{ children are girls}\]

Part A

Let us calculate \(P(G_n|E_1)\):

\[P(G_n|E_1) = \frac{P(G_n \cap E_1)}{P(E_1)}\] \[P(E_1) = P(E_1 \cap G_0) + P(E_1 \cap G_1) + P(E_1 \cap G_2) + .... + P(E_1 \cap G_n)\] \[P(E_1 \cap G_i) = P(E_1|G_i)P(G_i)\] \[P(E_1 \cap G_0) = P(E_1|G_0)P(G_0) = 0\]

Then,

\[P(E_1 \cap G_1) = P(E_1|G_1)P(G_1)\]

As we know that atleast one of the children is a girl:

\[P(E_1|G_1) = 1\]

Probability of having a single girl child considering all the combinations possible:

\[P(G_1) = \frac{1}{2^n}*{}^n \mathrm{ C }_1\]

So,

\[P(E_1 \cap G_1) = \frac{1}{2^n}*{}^n \mathrm{ C }_1\]

The above can be generalized as following:

\[P(E_1 \cap G_i) = \frac{1}{2^n}*{}^n \mathrm{ C }_i\]

So,

\[P(E_1) = \frac{1}{2^n}*\sum_{i=1}^n {}^n \mathrm{ C }_i\]

Using binomial expansion for \((1 + x)^n\) we find that the summation is equal to \(2^{n} - 1\)

\[P(E_1) = \frac{2^n - 1}{2^n}\] \[P(G_n|E_1) = \frac{P(G_n \cap E_1)}{P(E_1)}\] \[\begin{aligned} P(G_n|E_1) &= \frac{\frac{1}{2^n}*{}^n \mathrm{ C }_n}{\frac{2^n - 1}{2^n}} \end{aligned}\] \[\boxed{P(G_n|E_1) = \frac{1}{2^n - 1}}\]

Part B

Let us calculate \(P(G_n|E_2)\):

\[P(G_n|E_2) = \frac{P(G_n \cap E_2)}{P(E_2)}\] \[P(E_2) = P(E_2 \cap G_0) + P(E_2 \cap G_1) + P(E_2 \cap G_2) + .... + P(E_2 \cap G_n)\] \[P(E_2 \cap G_i) = P(E_2|G_i)P(G_i)\] \[P(E_2 \cap G_0) = P(E_2|G_0)P(G_0) = 0\]

Then,

\[P(E_2 \cap G_1) = P(E_2|G_1)P(G_1)\]

As we pick the child randomly and turns out to be a girl:

\[P(E_2|G_1) = \frac{1}{n}\]

Probability of having a single girl child considering all the combinations possible:

\[P(G_1) = \frac{1}{2^n}*{}^n \mathrm{ C }_1\]

So,

\[P(E_2 \cap G_1) = \frac{1}{2^n}*\frac{1}{n}*{}^n \mathrm{ C }_1\]

The above can be generalized as following:

\[P(E_2 \cap G_i) = \frac{1}{2^n}*\frac{i}{n}*{}^n \mathrm{ C }_i\]

So,

\[P(E_2) = \frac{1}{2^n}*\frac{1}{n}*\sum_{i=1}^n i*{}^n \mathrm{ C }_i\]

Using binomial expansion for \((1 + x)^n\) we find that the summation is equal to \(n*2^{n-1}\)

\[P(E_2) = \frac{1}{2}\] \[P(G_n|E_2) = \frac{P(G_n \cap E_2)}{P(E_2)}\] \[\begin{aligned} P(G_n|E_2) &= \frac{\frac{1}{2^n}*\frac{n}{n}*{}^n \mathrm{ C }_n}{\frac{1}{2}} \end{aligned}\] \[\boxed{P(G_n|E_2) = \frac{1}{2^{n-1}}}\]

Part C

Let us calculate \(P(G_n|E_3)\):

\[P(G_n|E_3) = \frac{P(G_n \cap E_3)}{P(E_3)}\] \[P(E_3) = P(E_3 \cap G_0) + P(E_3 \cap G_1) + P(E_3 \cap G_2) + .... + P(E_3 \cap G_n)\] \[P(E_3 \cap G_i) = P(E_3|G_i)P(G_i)\] \[P(E_3 \cap G_0) = P(E_3|G_0)P(G_0) = 0\]

Then,

\[P(E_3 \cap G_1) = P(E_3|G_1)P(G_1)\]

To calculate \(P(E_3|G_1)\) we know that the probability that atleast one girl is named \(Lilia\) is same is the complement of the probability that no girl is named \(Lilia\).

\[P(E_3|G_1) = 1 - P({E_3}^c|G_1)\] \[P({E_3}^c|G_1) = (1 - \alpha)^1\] \[P(E_3|G_1) = 1 - (1 - \alpha)^1\]

Probability of having a single girl child considering all the combinations possible:

\[P(G_1) = \frac{1}{2^n}*{}^n \mathrm{ C }_1\]

So,

\[P(E_3 \cap G_1) = \big[1 - (1 - \alpha)\big]*\frac{1}{2^n}*{}^n \mathrm{ C }_1\]

The above can be generalized as following:

\[P(E_3 \cap G_i) = \big[1 - (1 - \alpha)^i\big]*\frac{1}{2^n}*{}^n \mathrm{ C }_i\]

So,

\[P(E_3) = \frac{1}{2^n}*\sum_{i=1}^n \big[1 - (1 - \alpha)^i\big]*{}^n \mathrm{ C }_i\] \[P(E_3) = \frac{1}{2^n}* \Big[ \sum_{i=1}^n {}^n \mathrm{ C }_i - \sum_{i=1}^n (1 - \alpha)^i*{}^n \mathrm{ C }_i \Big]\]

Using binomial expansion for \((1 + x)^n\) we find that:

\[\sum_{i=1}^n {}^n \mathrm{ C }_i = 2^n - 1\] \[\sum_{i=1}^n (1 - \alpha)^i*{}^n \mathrm{ C }_i = (2 - \alpha)^n - 1\] \[P(E_3) = \frac{1}{2^n}*\big[ 2^n - 1 - ((2 - \alpha)^n - 1) \big]\] \[P(E_3) = \frac{1}{2^n}*\big[ 2^n - (2 - \alpha)^n \big]\] \[P(G_n|E_3) = \frac{P(G_n \cap E_3)}{P(E_3)}\] \[P(G_n \cap E_3) = \big[1 - (1 - \alpha)^n\big]*\frac{1}{2^n}*{}^n \mathrm{ C }_n\] \[\begin{aligned} P(G_n|E_3) &= \frac{\big[1 - (1 - \alpha)^n\big]*\frac{1}{2^n}*{}^n \mathrm{ C }_n}{\frac{1}{2^n}*\big[ 2^n - (2 - \alpha)^n \big]} \end{aligned}\] \[\boxed{P(G_n|E_3) = \frac{1 - (1 - \alpha)^n}{2^n - (2 - \alpha)^n}}\]

References

1. Probability course chapter 1 problems (Problem 37 & 39 & 38)
2. Finding the sum: \(\sum_{i=1}^n i*{}^n \mathrm{ C }_i\)