Probability with flipping a coin 6 times I have a fair coin and I flip it 6 times, here are the events:


*

*$Pr(A)$ = "the coin comes up heads at least 4 times"

*$Pr(B)$ = "the number of heads is equal to the number of tails"

*$Pr(C)$ = "there are at least 4 consecutive heads"


I am trying to determine the following: $Pr(A)$, $Pr(B)$, $Pr(C)$, $Pr(A|B)$, and $Pr(C|A)$
For $Pr(A)$ I did a sum of the binomial probability formula like this:
$Pr(A) = SUM k=4 to n= 6 C(n, k) (1/2)^k (1/2)^n-k = 11/32$
I have also manually wrote out all 64 combinations and counted them my answer is the same.
For $Pr(B)$ I did the same thing as I did for $Pr(A)$ but instead of a sum I did
$Pr(B) = C(6, 3) (1/2)^3 (1/2)^3 = 5/16$
Which again when I manually count it I get the same answer.
For $Pr(C)$, again jut counting I got $8/64$ $(1/8)$.
Not sure how to determine $Pr(A|B)$ and $Pr(C|A)$
Can someone point me in the right direction to determine this correctly without relying on listing all the combinations and counting?
thanks
 A: Just to elaborate this more into an answer:
For Pr(A), it is likely easy to break this down to the cases of 4,5, or 6 heads coming up where you could use the binomial distribution for each and then combine the results as I mentioned in the comment there are 6 cases of 5 heads, and 1 case of 6 heads to add in the end that you missed.
For Pr(B), consider that with 6 flips, this is 3 heads and 3 tails. Thus, you could use the binomial distribution in this case to pick which 3 places would you get heads or which 3 would you get tails, whichever way you want to see it.
For Pr(C), consider how this is a repeat of A in a sense since there are cases of 4,5, and 6 heads that could be valid cases. So, now the heads being consecutive makes for a block of 4 in a row, so the $HHHH$ combination if it is only 4 heads, means that there could be 0,1,2 or tails at the start and the others at the end of the 6 flips which is listing out this group initially. In the case of 5 heads there are a couple of different groups since there can be the $HTHHHH$ case in addition to 5 heads in a row as well as the reverse of these. The key here is to get things counted appropriately which is where a divide and conquer idea can work.
For Pr(A|B), note that these conflict as getting 4 heads given equal heads and tails so this would be 0.
For Pr(C|A), this would require looking into what relationship of getting at least 4 heads to getting 4 heads in consecutive fashion.
