On leap years and conditional probability Edit
(Copying this from the end of my post and pasting it at the start because people are still voting to close this.)
Before anyone flags this as a duplicate, I understand that the following exact problem has been posted before, but my post is about why my logic (for the second question below) is wrong (and not about what the correct answer should be).

Problem
A year in the 2020s (i.e. from 2020 to 2029 inclusive) is selected uniformly at random, a month is selected uniformly at random from that year and a day is selected uniformly at random from that month.

*

*What is the probability that the day is the 29th of February?

*Given that the day is the 29th, what is the probability that the month is February?

My idea for Question 1
The first question is very clear-cut.
First, we observe that only leap years have a "29th of February" and out of the entire decade, only 3 are leap years (i.e. 2020, 2024 and 2028), so we have a $\frac 3 {10}$ chance of picking a leap year.
By the same logic, we have a $\frac 1 {12}$ chance of picking February out of the leap year we have chosen and we have a further $\frac 1 {29}$ chance of picking the 29th of February from the February of the leap year we have chosen.
Thus, $\mathbb{P}(\mathrm{29th\ of\ February}) = (\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) = \frac 1 {1160}$.
Okay. Easy.
My idea for Question 2
Now, I understand that the second question is on conditional probability and Bayes' Theorem should be invoked. However, smart me thought I saw a short-cut (obviously incorrectly) and I have been wondering why my logic (as explained below) is wrong.
Since we are "given that the day is the 29th", I thought that we need only pick a year and a month. In other words, once we have a year and a month set, the probability that the day is the 29th should be $1$ right?
Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = (\frac 3 {10})(\frac 1 {12}) = \frac 1 {40}$.
Again. Easy. Or so I thought.
Suggested solution for Question 2
From the solution given (as explained below), it seems I have oversimplified Question 2. According to my professor, we need to break the question down much more than I had.
First, we must observe that, out of the entire decade (i.e. 120 months), we have 3 months with 29 days, 40 months with 30 days (since 4 months in a year have 30 days) and 70 months with 31 days (since 7 months in a year have 31 days). We do not count the "normal Februarys (i.e. the Februarys with 28 days)" as it is obvious that we cannot possibly have a 29th day from those months.
Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = \frac {(\frac 3 {120})(\frac 1 {29})} {(\frac 3 {120})(\frac 1 {29}) + (\frac {40} {120})(\frac 1 {30}) + (\frac {70} {120})(\frac 1 {31})} = \frac {279} {9965}$.

While I perfectly understand my professor's logic for the second question, I cannot help but wonder why my idea is flawed. Any intuitive explanations will be greatly appreciated!

For (1) you have a probability of 29th February chosen of $(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) = \frac 1 {1160}$
The equivalent probability of 29th March is $(\frac {10} {10})(\frac 1 {12})(\frac 1 {31})$ and there are $7$ months with $31$ days
The equivalent probability of 29th April is $(\frac {10} {10})(\frac 1 {12})(\frac 1 {30})$ and there are $4$ months with $30$ days
So that makes the probability of February given a 29th was chosen:
$$\dfrac{(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) }{1(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) +7(\frac {10} {10})(\frac 1 {12})(\frac 1 {31})+4(\frac {10} {10})(\frac 1 {12})(\frac 1 {30}) }= \dfrac{\frac 3 {29} }{\frac 3 {29}  +\frac{70}{31}+\frac {40} {30}} =\dfrac{279}{9965}$$ which is the suggested solution.
It is about $0.028$, slightly more than your $\frac1{40}=0.025$.  The reason is that because leap-year Februaries are shorter than Marches or Aprils, choosing a 29th is slightly more likely in such Februaries, so the reverse conditionality takes this into account.
