Why am I getting an incorrect answer from Bayes probability rule? The original question follows:



*If a [fair] coin is flipped 16 times, what is the probability of landing exactly 5 heads given that at least 3 of the flips landed heads?


Approach 1- Bayes Rule
My mistake is probably related to **.
I tried to apply Bayes rule.
$$ P( \space A|B \space ) = \frac{P(\space B|A\space) * P(\space A\space)}{P(\space B \space)}$$
A = exactly 5 heads.
B = 3 heads.  **Do I need $_{16}C_3 \space +\space _{16}C_4\space +\space 
 _{16}C_5\space ..._{16}C_{16} $ instead?
Hmmm, do I need to factor in that the three heads must occur in the first 14 flips?  Seems like that's not sufficient
$P( \space B|A\space) = 1$, because to get 5 heads, we first must get 3 heads.
$P(\space A\space) = _{16}C_5$
$P(\space B\space) = _{16}C_3$
Then $$ P( A|B) = \frac{_{16}C_5*(\frac12)^{16}}{_{16}C_3*(\frac12)^{16}}$$
Cancelling, this becomes $$\frac{13*12}{5*4}$$
Which is obviously incorrect.
Approach 2
So then I tried by each case.
$\begin{array}{|l|l|} \hline
\text{Description}&\text{Related Formula}\\\hline
\text{Begins with 3 tosses including exactly 3 heads}&_{3}C_3 *\space _{13}C_2 * (\frac12)^{16}\\
\text{Begins with 4 tosses including exactly 3 heads }&_{4}C_3 *\space _{12}C_2* (\frac12)^{16}\\ \vdots &\vdots\\
\text{Begins with 14 tosses including exactly 3 heads}&_{14}C_{3} *\space _{2}C_2* (\frac12)^{16}\\\hline
\end{array}$
Which sums to
$$\sum_{i=1} ^{12} \space _{i+2}C_3\space * _{14-i}C_2*(\frac12)^{16}$$
Since each row in the table or term in the sum is mutually exclusive I can simply add all the probabilities.  The 2nd approach seems better.  Is it correct?  And if so please explain why Bayes rule fails.
 A: 
And if so please explain why Bayes rule fails.

You were just not using the right events.  You were using "exactly 3" when you needed "at least 3".

The 2nd approach seems better. Is it correct?

It is better, but certainly not correct. You do not need the condition: "begins with three heads," or anything like that.

Hmmm, do I need to factor in that the three heads must occur in the first 14 flips?

No, because the condition is just that there are at least three heads somewhere among the sixteen tosses.
Don't add unnecessary constraints.



If a [fair] coin is flipped 16 times, what is the probability of landing exactly 5 heads given that at least three of the flips land heads?

Do I need $\def\C#1#2{{^{#1}\mathrm C_{#2}}}\C{16}3 +\C{16}4+ \C{16}3 \ldots \C{16}{16}$ instead?  See note**.

Yes, because the condition is that it is at least, rather than exactly.
Although this is also $2^{16}-\C{16}0-\C{16}1-\C{16}2$, by the Binomial Expansion Theorem.

Let $H$ be the count for heads among sixteen tosses of a fair coin:
$$\begin{align}\mathsf P(H=5\mid H\geq 3) & = \dfrac{\mathsf P(H=5)}{1-\mathsf P(H\lt 3)}\\[2ex]&=\dfrac{\C{16}5}{2^{16}-\C{16}0-\C{16}1-\C{16}2}\cdot\color{gray}{\require{cancel}\dfrac{\cancel{1/2^{16}}}{\cancel{1/2^{16}}}}\end{align}$$

$\small\textbf{Note: }\{H=5\}=\{H=5\}\cap\{H\geq 3\}$
** $^{n}C_{r}$ is simply an alternate notation for $_{n}C_{r}$.
