Probability of an odd number of heads when flipping a fair coin 3 times. if I flip a fair coin $3$ times, what is the probability that  the coin comes up heads an odd number of times.
any help please. 
I understand the probability(A=the coin comes up heads an odd number of times)=1/2. probability (B=the coin comes up tails an odd number of times)=1/2 but this got me confusing probability(A|B)? 
 A: There's only three tosses so a brute force solution seems the simplest and most intuitive to me:
       odd?
H H H  yes
H H T   no
H T H   no
H T T  yes
T H H   no
T H T  yes
T T H  yes
T T T   no

4 out of 8 possibilities: 50% chance
A: Possible odd numbers: 1, 3
By the sum rule: P(Odd) = P(1) + P(3)
Use the binomial probability formula:
$P(1) = {3 \choose 1}p^1(1-p)^2$. Note that $p = 1-p = 1/2$ because the coin is fair. So $P(1) = 3p^3 = 3/8$.
Similarly, $P(3) = {3 \choose 3}p^3(1-p)^0 = 1p^3 = 1/8$.
So P(Odd) = $1/8 + 3/8 = 4/8 = 1/2$.
A: The coins are tossed onto a glass-topped coffee table. I am under the table, looking up (don't  ask). 
The person tossing sees an odd number of heads if and only if I see an even number of heads. But by symmetry odd number of heads is just as likely as odd number of tails.
Thus odd number of heads and even number of heads are equally likely, and  the probability of an odd number of heads is $\frac{1}{2}$.
A: Macavity's comment and André's answer use a "global" symmetry that requires the total number of flips to be odd. Interestingly, though, the probability is also $\frac12$ if the total number of flips is even, and this is due to a more general "local" symmetry: The last coin flipped decides whether the total number of heads is odd or even, and the last coin is equally likely to come up heads or tails.
This is related to the vanishing of the alternating sum of binomial coefficients,
$$
\sum_{k=0}^n(-1)^k\binom nk=0\;.
$$
For odd $n$, this can be seen explicitly using $\binom nk=\binom n{n-k}$, which corresponds to the "global" symmetry. For general $n$, it can be seen either from the fact that this is the binomial expansion of $(1-1)^n=0^n$, or from a combinatorial argument that's essentially the same as the last coin argument above: The sum counts the excess of even-sized over odd-sized subsets of a set of $n$ elements, which you get from the subsets of the first $n-1$ elements by either adding the $n$-th element or not, which yields one odd-sized and one even-sized subset in each case.
A: Hint: How many results can you get total from flipping a coin 3 times? Think of them as triples, e.g. THH for tails, then heads, then heads, or HTH for heads, then tails, then heads. (Since it's fair, each such result has equal probability.)
Then, how many results have 1 and only 1 coming up heads? How many results have all three coming up heads?
