Conditional Probabilities, babies using Bayes 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?
$$P(\text{Boy}\,|\,\text{Mother}),\, P(\text{Boy}),\, P(\text{Mother})$$

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
