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In a Leap year a month is selected at random and a day is selected at random and found that its fifth Friday. What is the Probability that selected month has $30$ days.

My try: Let $A$ be an event of day chosen is fifth friday

$M_{30}$ be an event of chosen month having $30$ days

$M_{29}$ be an event of chosen month having $29$ days

$M_{31}$ be an event of chosen month having $31$ days

$P(M_{30})=\frac{4}{12}$

$P(M_{29})=\frac{1}{12}$

$P(M_{31})=\frac{7}{12}$

we need to find

$$P\left(M_{30}/A\right)$$

By Bayes theorem we have

$$P\left(M_{30}/A\right)=\frac{P\left(A/M_{30}\right)P(M_{30})}{\sum P(A)}$$

but how to find $P\left(A/M_{30}\right)$?

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  • $\begingroup$ Assuming you are working with Gregorian calendar (which is a pretty reasonable assumption, unless you want to work with Julian calendar for e.g. astronomical reasons), use its minimal period of 400 years to calculate $\mathbf{P}(A\cap M_{30})$. $\endgroup$ – user10354138 Oct 14 '18 at 15:24
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I think you are on the right track.

Now, I'd define $B$ as the event : "the month has a fifth friday".

Then $$ \begin{align}P(A \mid M_i) &= P( A, B \mid M_i) + P( A, B^c \mid M_i) \\ &= P( A, B \mid M_i) \\ &= P( A \mid B ,M_i) \, P(B \mid M_i) \end{align}$$

Can you go on from here?

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