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You know that family has two children and at least one of these is a daughter. Now suppose that you ring the doorbell and a daughter opens the door. Assuming that each of the two children answer the doorbell with equal probability what is the probability you now assign to the family having two daughters?

A: family has two daughters; B : family has one daughter; C : daughter opens the door

$P(C/A) = 1$

$P(C/B) = \frac 1 2$

$P(A/C) = \frac 1 {1+ \frac 1 2} = \frac 2 3 $

Is this correct ?

gandalf61
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Anjali
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  • Your working is a little unclear, and I'm not sure how you got $1/1+1/2=2/3$, but consider that P(2 daughters) = P(2 daughters | known child answers) $\cdot$ P(known child answers) + P(2 daughters | unknown child answers) $\cdot$ P(unknown child answers) $= 1/2\cdot 1/2+1\cdot 1/2=3/4$. – Tom Feb 02 '23 at 11:48
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    What you wrote is hard to follow. First of all, $\frac 11+\frac 12$ is not $\frac 23$. Secondly, you never seem to consider the a priori probabilities of the events $A,B$. Those surely matter...seeing the daughter at the door allows you to revise your estimate, but you need to start somewhere. – lulu Feb 02 '23 at 11:48
  • Does this answer your question? Probability that both 2 already born children are girls given that one of them is a girl – Ethan Bolker Feb 02 '23 at 12:15
  • Of course , in those exercises it is assumed that the probability for each child to be a girl is 1/2 , whether this is a realistic assumption or not. With this assumption, this is a simple exercise about Bayes' theorem. – Peter Feb 02 '23 at 12:19
  • @EthanBolker I don't think the question you linked to is an equivalent question to this one. – gandalf61 Feb 02 '23 at 12:30

1 Answers1

1

It looks like you are trying to apply Bayes' theorem which tells us that

$\displaystyle P(A|C) = \frac {P(C|A)P(A)}{P(C)}$

where event A is "family has two daughters given that the family has two children and at least one daughter" and event C is "a daughter answers the door given that the family has two children and at least one daughter".

You know that $P(C|A)=1$ and $P(A) = \frac 1 3$. If $\lnot A$ (which I think you are calling B) then the family has one son and one daughter, so $P(C|\lnot A) = \frac 1 2$ (this is where we use the information that one of the children will answer the door, and each child is equally likely to answer the door). So we can find $P(C)$ as follows

$\displaystyle P(C) = P(C|A)P(A) + P(C|\lnot A)P(\lnot A) = \frac 1 3 + \frac 1 3 = \frac 2 3$

which gives us

$\displaystyle P(A|C) = \frac 1 3 \times \frac 3 2 = \frac 1 2$

Another way to see this is as follows. There are two scenarios in which a daughter answers the door:

  1. The family has two daughters and one of them answers the door - probability $\frac 1 3$.
  2. The family has one son and one daughter and the daughter answers the door - probability $\frac 2 3 \times \frac 1 2 = \frac 1 3$.

Since these two scenarios are equally likely, the probability that we have scenario 1 is $\frac 1 2$.

gandalf61
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