# Independent Events and Coin Flips

We flip a fair coin three times. For $$k=1,2,3$$ let $$A_k$$ denote the event that there are an even number of heads within the first $$k$$ coin flips.

a) Are $$A_2$$ and $$A_3$$ independent?

b) Are $$A_1, A_2, A_3$$ (mutually) independent?

Note that our sample space is $$\Omega=\{ (HHH),(HHT),(HTT),(HTH),(THH),(TTT),(TTH),(THT)\}$$

For a) and b), we have by brute calculation, that $$P(A_1)=P(A_2)=P(A_3)=1/2,$$ also $$P(A_2\cap A_3)=1/4,$$ and $$P(A_1\cap A_2 \cap A_3)=P(A_1)P(A_2)P(A_3)=1/8$$, so that $$A_1,A_2$$ and $$A_3$$ are mutually independent, hence so are $$A_2,A_3$$.

• $P(A_2\cap A_3)\neq \frac 12$. Indeed, the events are independent, so the probability of the intersection is $\frac 14$. – lulu Feb 15 at 12:26
• @lulu Sorry, made a typo. – G the Stackman Feb 15 at 12:27

This is true for any $$A_i,A_j$$ with $$i.
Indeed, it is clear that $$P(A_n)=\frac 12$$ for all $$n$$.
To analyze $$P(A_i\cap A_j)$$ let $$B_{j,i}$$ be the event that there are evenly many Heads tossed between the $$(i+1)^{st}$$ and the $$j^{th}$$ toss. Of course $$P(B_{j,i})=\frac 12$$. Then the event $$A_i\cap A_j=B_{j,i}\cap A_i$$ And it is clear that $$B_{j,i}$$ and $$A_i$$ are independent.