Sunny Days in Mathtown

On day 1, $P(sunshine) = p$ .

On day 2, $P(sunshine) = p \times \frac{9}{10} + (1-p) \times \frac{3}{10} = p$ .

Solving, $p = \frac{3}{4}$ and $3 + 4 = 7$ .

The probability of it raining a particular day is $1-p$ and the probability of the weather being sunny after a rainy day is $1-\frac{7}{10}=\frac{3}{10}$ . It can be sunny after a rainy day or a sunny day so the probability of a particular day being sunny, which is p, can be written as:

$p=p\times \frac{9}{10}+(1-p)\times \frac{3}{10}$

Solving for p will give $p=\frac{3}{4}$ , hence the answer is $3+4=7$

The probability that is sunny on March 1st is $p$ , while the probability of being rainy on March 1st can be expressed as $1-p$ .

The probability of it being sunny on March 2nd can be calculated by summing up the two different weather conditions we'd had on March 1st as follows:

$\frac{9}{10} \times p + \frac{3}{10} \times (1-p)$

It is also stated in the problem definition that the probability of being sunny on any given day of March (e.g March 2nd) is $p$ . Therefore we can write the following equation:

$p=\frac{9}{10} \times p + \frac{3}{10} \times (1-p)$

By solving the simple equation above, we'll have $p=\frac{3}{4}$ . Thus the final answer would be $7$ .

given "the probability of a day being sunny on any given day during the month is identically p" consider day 2

probability that day 2 is sunny= day 1 sunny and day 2 sunny + day 1 rainy and day 2 sunny

p= p . (9/10) + (1-p)(1-7/10)

p=3/4