Conditional probability on 3 coin tosses I'm having trouble with a problem which I think I already found a way to answer, but I can't really identify where is the error on an alternate solution I was trying to develop.
This was taken from page 31 of Mathematical Statistics and Data Analysis, John Rice.
A fair coin is tossed three times.
a. What is the probability of two or more heads given that there was at least onehead?
b. What is the probability given that there was at least one tail? 
(a) I took $N$ as the number of heads, so: $P\left( {N \ge 2|N \ge 1} \right)$, which using Bayes' Rule, got me  $${{P\left( {N \ge 1|N \ge 2} \right)P\left( {N \ge 2} \right)} \over {P\left( {N \ge 1} \right)}} = {{1.{4 \over 8}} \over {{7 \over 8}}} = {4 \over 7}$$  which seems to be correct according to the books final answer.  
(b) Here I'm at a loss. It asks me for $P\left( {N \ge 2|N \le 2} \right)$, which is only possible if I search specifically for $P\left( {N = 2} \right) = {3 \over 7}$, which is the correct answer. The problem is I tried to "check" if the Bayes' Rule would take me there too (as it should), and somewhere along the way I am getting lost.  
$P\left( {N \ge 2|N \le 2} \right) = {{P\left( {N \le 2|N \ge 2} \right)P\left( {N \ge 2} \right)} \over {P\left( {N \le 2} \right)}}$$  
$$ = {{{3 \over 8}{4 \over 8}} \over {{7 \over 8}}} = {3 \over {14}}$
If anyone can help just by pointing the error, the direction, I'd really appreciate.
Cheers!
 A: You have made the common error for dealing with conditional probabilities.
Firstly, $\Bbb P(N=2) \neq \frac 37$. You can have HHT, HTH, or THH: $3$ favourable outcomes out of $8$ possible outcomes, so it should be $\Bbb P(N=2) = \frac 38$
Or if you prefer binomial, then
$$\Bbb P(N=2) = C^3_2 \bigg(\frac 12 \bigg)^2 \bigg(\frac 12 \bigg)^1 = \frac 38$$
Next, you have correctly identified that $N \geq 2 | N \leq 2$ is equivalent to $N = 2$. However, it should be
$$\Bbb P(N \geq 2 | N \leq 2) = \Bbb P(N = 2 | N \leq 2)$$
When you modify the event $N \geq 2$ based on the conditioned event, you still have to keep the condition. Thus, using your approach, you should obtain
$$\Bbb P(N \geq 2 | N \leq 2) = \Bbb P(N = 2 | N \leq 2) = \frac{\Bbb P(N=2 \cap N \leq 2)}{\Bbb P(N \leq 2)} = \frac{\Bbb P(N=2)}{\Bbb P(N \leq 2)} = \frac{\frac 38}{\frac 78} = \frac 37$$
I assume you made a similar mistake in your "other" method.

Basically, people tend to mix up the $\cap$ and the $|$. The difference is:
$$\Bbb P(N \geq 2 | N \leq 2) = \Bbb P(N = 2 | N \leq 2)$$
whereas
$$\Bbb P(N \geq 2 \cap N \leq 2) = \Bbb P(N = 2 )$$
