If the probability of heads when tossing a coin is p. This coin is tossed n times. If H is the event of getting a Head on the first toss and $F_k $ is the event exactly k Heads are tossed, for which pairs of integers (n,k) are H and $F_k $ independent,

I can't really see how these events will ever be independent? Given the first toss is a head it will always affect the event of an exact number of total heads? But the question suggests finding these pairs of integers? any help?

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    $\begingroup$ Try finding both $P(H\cap F_k)$ and $P(H)P(F_k)$. Then find which values of $n,k,p$ wll make these equal. I think you will find it is possible for some values. $\endgroup$ – Mick A Feb 15 '15 at 22:01
  • $\begingroup$ I get $k=pn$. Either way, doesn't that give infinite possible integer solutions for $k,n$ if $p$ is rational? $\endgroup$ – Mick A Feb 15 '15 at 22:12

By definition two events A and B are independent if $P(A\cap B)=P(A)P(B).$

In this case $P(A)=P(H)=$Probability of getting heads on first try and

$P(B)=P(F_k)=$Probability of getting exactly k heads. We know that $P(A)=p$ and we also know that $P(B)={n \choose k}p^k(1-p)^{n-k}$ so $P(A)P(B)=p{n \choose k}p^k(1-p)^{n-k}$.

$P(A\cap B)$ is the probability of getting heads on the first toss and then exactly $k-1$ more heads so $P(A\cap B)=p{n-1 \choose k-1}p^{k-1}(1-p)^{n-k}$.

Setting the two equal to each other yields ${(n-1)! \over (k-1)!(n-k)!} p^k(1-p)^{n-k}={(n)! \over (k)!(n-k)!}p^{k+1}(1-p)^{n-k}$ which upon simplification yields $k=np$ so for instance if $p={1 \over 2}$ and $n=2$ then $H$ and $F_k$ are independent.

  • $\begingroup$ I think $(1-p)^{n-(k-1)}$ should be $(1-p)^{n-k}$ - we need $n-k$ tails for event $A\cap B$. $\endgroup$ – Mick A Feb 15 '15 at 22:26
  • $\begingroup$ True. Thanks I made the edits $\endgroup$ – Ryan Feb 15 '15 at 22:35
  • $\begingroup$ Technically, $H$ is an event as described above and therefore a set, not a value. You mean to write $A=H$ and $P(A)=P(H) =$ Probability of..., Similarly for $B=F_k$. $\endgroup$ – JMoravitz Feb 15 '15 at 22:38
  • $\begingroup$ Hi @Ryan Just in the example at the end: it's only independent when $k=1$. $\endgroup$ – Mick A Feb 15 '15 at 22:43

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