# Probability that a family contains atleast one male and atleast one female child

In a colony all families have at least one child.The probability that a randomly chosen family from this colony has exactly $k$ children is $(0.5)^k;k=1,2,...$ A child is either a male or a female with equal probability.What will be the probability that such a family consists of atleast one male and atleast one female child?

Well,I started the problem in this way,may be I am wrong.

$A_i$:The $ith$ family has no male and female child. So, we have to compute $P(\bigcap_{i=1}^\infty A_i)^c$

$P(\bigcap_{i=1}^\infty A_i)^c=1-P(\bigcup_{i=1}^\infty A_i)=1-(\sum P[A_i]-\sum(P(A_{i1}\bigcap A_{i2})+...+P(\bigcap A_i))$

Now I am unable to compute the probabilites. May be my approach is not right.Can anyone please give some hints about this problem?

• $A_i$ seems to say this particular child is neither male nor female! Why don't you compute the chance that a family has all girls, then go on from there? If that's too hard, compute the chance that a family has $k$ children and all of them are girls. – whuber Jun 2 '16 at 15:03
• So Priyanka did you get admission through JAM? XD – Ronald Aug 17 '18 at 15:40

Let's break this down into three possibilities:

1. Only male children
2. Only female children
3. A mixture of male and female children

The three groups sum to 1 (because all families have at least one child), and so if we compute the probability of the first two groups, we can subtract that from one and get our answer. This is even easier because the two are symmetric--the probability of only having male children is the same as the probability of only having female children.

So how would we get the probability of only having sons?

Start with the probability of having a family with exactly one son (and zero daughters). The chance that the family size ($F$) is 1 is:

$P(F=k)=0.5^k \to P(F=1)=0.5$

The probability that a family of size 1 has only sons ($S=F$) is:

$P(S=k|F=k)=0.5^k \to P(S=1|F=1)=0.5$

But we want an unconditional probability, so

$P(S=1,F=1)=P(S=1|F=1)P(F=1)=0.25$.

Back out to the formulation with $k$, sum over all $k$, and we have half of the complement of our answer.

• What do you mean we need unconditional probability. Did you just make unconditional by just multiplying with $P(F=1)$ – Ronald Aug 17 '18 at 17:39
• @Damn1o1, the final calculation is for all family sizes, and so we can't use a number that's conditioned on the family size. I used the Law of Total Probability there. – Matthew Graves Aug 25 '18 at 3:15