The probability of rain in a given week is $2/7$. How do you find probability it will rain for exactly 4 days ?
I used the formula as $({}^7C_4) \cdot (0.28)^4 \cdot (0.72)^3$ here, where the probability of rain $= 2/7 = 0.28$ and the probability of no rain = $(1-0.28) = 0.72$
Not sure my math is right or wrong. Need suggestion to be sure.
It is like finding the number of binary numbers with seven digits, four of them being ones.
As you have guessed, it is a case where the convenient model is Binomial distribution $B(n,p)=B(7,p)$ in which you are interest by the event where the random variable "number of rainy days = X" is equal to 4. We know that, with this model:
$$\tag{1}P(X=4)=\binom{7}{4}p^7(1-p)^4,$$
but with which $p$ ? Here is where we disagree. I don't have the same $p$ as you:
For me, the value of $p$ is determined by constraint:
$$\text{Probability of at least a rainy day in the week} \ = 1-\underbrace{(1-p)^7}_{\text{not a single rainy day}}=2/7$$
$$\iff p=1-(5/7)^{1/7}=0.0469 \ \ \text{(not very simple !)}$$
It remains to plug this value of $p$ into (1) to obtain the result.