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 $A_2,A_3$.

  • $\begingroup$ $P(A_2\cap A_3)\neq \frac 12$. Indeed, the events are independent, so the probability of the intersection is $\frac 14$. $\endgroup$ – lulu Feb 15 '19 at 12:26
  • $\begingroup$ @lulu Sorry, made a typo. $\endgroup$ – G the Stackman Feb 15 '19 at 12:27

This is true for any $A_i,A_j$ with $i<j$.

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.


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