Conditional Probabilities, babies using Bayes

Question

There are $3$ boys and $x$ girls in a hopital (babies). A mother gives birth to a child but doesn't know it's gender. A midwife picks a random baby, given that she picked a boy, what is the probability it comes from the mother?

I used bayes, but I am struggling to specify (understand/make sense of) the probabilities. Can someone please explain what each of the following probabilities are?

$$P(\text{Boy}\,|\,\text{Mother}),\, P(\text{Boy}),\, P(\text{Mother})$$

Answer 1

With the new information, I made a correction. Also, I added a rather philosophical argument, at the end.

Firstly, the experiment is to take a baby from $4+x$ babies, at random.

$P(B)$ is the probability that a picked baby is a boy. Therefore,

$$P(B)=\frac{1}{2}\frac{4}{4+x}+\frac{1}{2}\frac{3}{4+x}=\frac{7}{2(4+x)}$$

$P(M)$ is the probability that a picked baby belongs to the mother. Therefore

$$P(M)=\frac{1}{4+x}$$

For $P(B|M)$, which is the probability that a picked baby is a boy, when we know that it belongs to the mother, we need to look at it from the mother point of view. She knows that it would be fifty-fifty. Therefore,

$$P(B|M)=\frac{1}{2}$$

Then you just need to use Bayes theorem to calculate

$$P(M|B)=\frac{1}{7}$$

Extension:

Bayesian approach is used when something cannot be measured, using frequentist approach. This problem is a good example. Assume we make the baby, that belongs to the mother, blue and we put a patch on it, so the gender is not observable. If we want to calculate $P(M|B)$, using frequencies, then we have to separate boys and do the following. Take one of the boys at random and show him to the mother. We need to do it many times and Mother would calculate the probability. But, in order to do such a thing, we need to know if the blue baby belongs to boys or not. So, the scenario is not well-defined and therefore, not accurately measurable. That is why we resort to Bayesian approach and calculate things that we can more reliably approximate, like the the $P(M)$, $P(B)$ and combine it with a mere guess of $P(B|M)=0.5$.

Answer 2

With the information given and reading "There are $3$ boys and $x$ girls in a hospital" as the position after the mother has given birth, I would have thought the probabilities were independent, i.e.

• $P(\text{mother's child picked})= \dfrac{1}{3+x}$
• $P(\text{boy picked})= \dfrac{3}{3+x}$
• $P(\text{mother's child picked and it is a boy})= \dfrac{3}{(3+x)^2}$
• $P(\text{mother's child picked}\mid \text{boy picked})= \dfrac{1}{3+x}$
• $P(\text{boy picked}\mid \text{mother's child picked})= \dfrac{3}{3+x}$

(Added later)

If the "There are $3$ boys and $x$ girls in a hospital" is the position before the mother gives birth, then I would have said

• $P(\text{mother's child is a boy})= \dfrac{1}{2}$
• $P(\text{mother's child picked})= \dfrac{1}{4+x}$
• $P(\text{mother's child picked} \mid \text{mother's child is a boy})= \dfrac{1}{4+x}$
• $P(\text{boy picked} \mid \text{mother's child is a boy})= \dfrac{4}{4+x}$
• $P(\text{boy picked} \mid \text{mother's child is a girl})= \dfrac{3}{4+x}$
• $P(\text{boy picked})=\dfrac12 \dfrac{4}{4+x} + \dfrac12 \dfrac{3}{4+x} = \dfrac{7}{2(4+x)}$
• $P(\text{mother's child picked and it is a boy}) = \dfrac12 \dfrac{1}{4+x} = \dfrac{1}{2(4+x)}$
• $P(\text{mother's child picked} \mid \text{boy picked}) = \dfrac{\frac{1}{2(4+x)}}{\frac{7}{2(4+x)}}= \dfrac{1}{7}$

which is the same as Cato's answer

Answer 3

P(handed her child | child handed over is boy) = P(handed her baby boy AND baby handed over was boy) / P( baby handed over was boy)

= $[\frac{1}{2} \times \frac{1}{x + 4}] / [ \frac{1}{2} \times \frac {3}{x + 4} + \frac{1}{2} \times \frac{4}{x + 4}$]

$= \frac{1}{7}$

I think it is independent of x, the number of female babies - the mother has already had a boy or a girl, so it is boy with probability 1/2, in which case there would be 4 equally likely circumstances she would be handed a boy, one of which is hers - if it was a girl she had, she can be shown 3 equally likely boys not hers.

Another way of looking at it is that she has given birth to half a boy, to add to the 3 boys $\frac{\frac{1}{2} }{3 \frac{1}{2}} = \frac{1}{7}$

So that's my answer, the chances the mother is looking at her baby is 1/7