Leap year probability 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 of the month is selected uniformly at
random from that month.
(a) What is the (exact) probability that the day is the 29th of February?
->I use 3/10 * 1/12 * 1/29 [not very sure]
(b) Given that the day is the 29th, what is the probability that the month is February?
For this one, I know I should use Baye's rule, but cannot proceed further. Is anyone willing to help?
 A: Your answer (a) is correct.
You can calculate the probability $P_{31}$ that the day you draw is the 29th of a month with exactly 31 days:
$$P_{31}= \frac{7}{12}×\frac{1}{31}.$$ [Duly note that every year has exactly 7 months with 31 days.]
And then you can calculate the probability $P_{30}$ that the day you draw is the 29th of a month with exactly 30 days:
$$P_{30}=\frac{4}{12}×\frac{1}{30}.$$
You already calculated above in (a) correctly that the probability $P_{29}$ that the day you draw is the 29th of February, 2020, 2024, or 2028.
$$P_{29}=\frac{3}{10}×\frac{1}{12} ×\frac{1}{29}.$$
Then the probability $P$ that you draw is the 29th of a month, satisfies
$$P = P_{31}+P_{30}+P_{29}.$$ And so, the probability that you the day you draw is the 29th of some February given that the day you drew is the 29th of some month, is $\frac{P_{29}}{P} = \frac{P_{29}}{P_{31}+P_{30}+P_{29}}.$ Check to make sure you see why this is.
Can you use this to work out the numbers and finish (b).
A: Your answer to a is correct. You must choose a leap year, and then you must choose February, and then you must choose the 29th.
For part (b), $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ Setting $A = $ "February" and $B = $ "29th", we find $P(A) = 1/12$, regardless of the year chosen.
Finding $P(B)$ is the sum of all the ways it can happen.  $$\frac{1}{10}\cdot \frac{1}{12} \cdot \frac{1}{31}$$ is the probability of getting January 29th, 2020. Likewise there are another $112$ possibilities. We just need to group them.  $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{1}{31} + \frac{1}{31} + \frac{1}{30} + \frac{1}{31} + ...]$$ has eleven terms, $7$ of which are $\frac{1}{31}$ and $4$ of which are $\frac{1}{30}$, giving us $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{7}{31} + \frac{4}{30}]$$ gives the probability for non-leap years. Add to that $$\frac{3}{10} \cdot \frac{1}{12} \cdot [\frac{7}{31} + \frac{4}{30} + \frac{1}{29}]$$ to get the total for $P(B)$.
