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
Best Answer
We can classify all of the possible "29th" dates as follows:
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 $\frac{279}{9965}$.