What is the probability of at least $2$ heads (not neccessarily consecutive) will appear when a coin is flipped $9$ times?
I know how to do this using Combination/Permutation with repetition, but this time around I want to solve by applying the Inclusion-Exclusion Principle, which I am unsure of how to do.
I know that I must use the inclusion-excluion formula: $$N(\overline A_1 \cap \overline A_2 \cap \cdots \cap \overline A_n) = N - S_1 + S_2 - S_3 + \cdots + (-1)^k S_k + \cdots +(-1)^n S_n$$
Where each $A_n$ is a coin toss resulting in Tails.
But what does $k$ represent in this problem, and how would I apply the formula?
That is equal to $1$ minus the probability of exactly one head and the probability of no heads. If a fair (unbiased) coin is flipped nine times, the probability of no heads (all tails) is $$\left(\frac{1}{2}\right)^9$$ The probability of exactly one head is $\left(\dfrac{1}{2}\right)^9$ times the number of different ways that a head could occur, $9$, so: $$9 \times \left(\frac{1}{2}\right)^9$$ The probability of at least two heads is: $$1- 9\cdot\left(\frac{1}{2}\right)^9-\left(\frac{1}{2}\right)^9= 1- 8\cdot\left(\frac{1}{2}\right)^9= 1-\left(\frac{1}{2}\right)^6$$