Conditional probability of selecting day in a months 
A year is chosen at random from  a set of numbers $\{ 2012,2013,\dots,2021 \}$. From the chosen year, a month is chosen at random and from the chosen months, a day is chosen at random. Given that the chosen day is the $29^{\mathrm{th}}$  of the months, the conditional probability that the chosen months is February equal to?

Probability of selecting a year from leap year as $\displaystyle \frac{1}{3}$ or probability of non-leap year as $\displaystyle \frac{1}{7}$
And probability of selecting a months from $12$ months as $\displaystyle \frac{1}{12}$
And selecting a day in a february month as $\displaystyle \frac{1}{28}$ or $\displaystyle \frac{1}{29}$ whether is is leap year or ñon leap year.
So required probability of selecting
a day in february months as $\displaystyle \frac{1}{3}\cdot \frac{1}{12}\cdot \frac{1}{29}+\frac{1}{7}\cdot \frac{1}{12}\cdot \frac{1}{29}$
But my answer is wrong, help me
 A: We can classify all of the possible "29th" dates as follows:

*

*Feb 29ths. The probability of selecting any given one of these (e.g. Feb. 29, 2020) is $\frac1{10}\frac1{12}\frac1{29}$, and there are 3 of them (2012, 2016, 2020).

*The 29th of a 31-day month. The probability of picking any given one of these (e.g. Jan. 29, 2020) is $\frac1{10}\frac1{12}\frac1{31}$, and there are $10\cdot7$ of them (in Jan, Mar, May, Jul, Aug, Oct, Dec of each year).

*The 29th of a 30-day month. The probability of picking any given one of these (e.g. Apr. 29, 2020) is $\frac1{10}\frac1{12}\frac1{30}$, and there are $10\cdot4$ of them (in Apr, Jun, Sep, Nov of each year).

Therefore, the probability that we select a 29th is $3\times\frac1{10}\frac1{12}\frac1{29}+70\times\frac1{10}\frac1{12}\frac1{31}+40\times\frac1{10}\frac1{12}\frac1{30}$.
The probability that we select a February 29th is just the first term: $3\times\frac1{10}\frac1{12}\frac1{29}$.
So given that a 29th was selected, the conditional probability that February was selected is the ratio of these:
$$\frac{3\times\frac1{10}\frac1{12}\frac1{29}}{3\times\frac1{10}\frac1{12}\frac1{29}+70\times\frac1{10}\frac1{12}\frac1{31}+40\times\frac1{10}\frac1{12}\frac1{30}}
$$
Notice that the factors of $\frac1{10}\frac1{12}$ all cancel out, so this simplifies to
$$\frac{\frac3{29}}{\frac3{29}+\frac{70}{31}+\frac{40}{30}}$$
which is equal to $\frac{279}{9965}$.
