Logical Error at Conditional Probability Question I am having a rather tough time wrapping my head around any possible logical fallacy in my solution to the following question as my answer is wrong:
1 percent of children have autism. A test for autism is developed such that 90% of autistic children are correctly identified as having autism but 3% of non-autistic children are mistakenly identified for having autism. A child is tested at 2 independent clinics. What is the probability that both clinics give the same diagnosis? 
My solution:
Find the probabilities of any individual being tested positive and being tested negative. Then, square them and add them together (two clinics). 
P(positive test) = (Has autism)(tested positive given has autism) + (Does not have autism)(tested positive given does not have autism)
Which would be: (0.01 * 0.9) + (0.99 * 0.03) = 0.0387
Similarly, the test for negative would be calculated to be 0.9613. Hence, my answer would be (0.0387)^2 + (0.9613)^2 = 0.9255
However, this is not the same answer as the solution. The solution suggests to do it in the way of P(Positive in one clinic & Positive in another clinic | Autistic) * P(A) + P(Negative in one clinic & Negative in the other clinic | Not autistic) * P(not Autistic) which would yield 0.94.
I don't quite understand why this is different from my solution though. Just where exactly am I making a logical error? 
EDIT: Revised; thanks Michael Hardy for correcting me.
 A: The second approach suggested ought to say this:
\begin{align}
& \Pr(\text{positive in one clinic & positive in another clinic} \mid \text{autistic})\cdot\Pr(\text{autistic}) \\
{} + {} & \Pr( \text{negative in one clinic & negative in the other clinic} \mid \text{autistic}) \cdot \Pr(\text{autistic}) \\
{} + {} & \Pr(\text{positive in one clinic & positive in the other clinic} \mid \text{not autistic}) \cdot \Pr(\text{not autistic}) \\
{} + {} & \Pr(\text{negative in one clinic & negative in the other clinic} \mid \text{not autistic}) \cdot \Pr(\text{not autistic})
\end{align}
Another problem: Your approach treats the two clinics' outcomes as independent.  They're not.  They are conditionally independent given the autism status. What is the conditional probability of a positive result at one clinic given a positive result at the other?  It is more than the unconditional probability of a positive test, because, given a positive test at one clinic, the probability of autism is then more than $0.01$.
$$
\Pr(\text{autism}\mid + ) = \frac{\Pr(\text{autism}) \cdot \Pr(+ \mid \text{autism})}{\Pr(+ )} = \frac{0.009}{0.009 + 0.0297 } \approx 0.232558.
$$
Another issue is this: We're testing the SAME person both times. That also interferes further with independence. For example, it is consistent with all of the information given in the statement of the problem that every patient either always tests positive at every clinic or always tests negative at every clinic. So even conditional independence is a stretch.
A: 
1 percent of children have autism. A test for autism is developed such that 90% of autistic children are correctly identified as having autism but 3% of non-autistic children are mistakenly identified for having autism. A child is tested at 2 independent clinics. What is the probability that both clinics give the same diagnosis? 

Step 1: Identify what is given.


*

*Let $A$ be the event that the child has autism.  $\mathsf P(A) = 0.01$

*Let $T_1, T_2$ be the events that the tests are positive at the two clinics, respectively.  $\mathsf P(T_n\mid A) = 0.90, \mathsf P(T_n\mid \neg A)=0.03$
Step 2: Identify what is asked.
$$\begin{align}\mathsf P(T_1=T_2)
 & = \mathsf P(T_1 = T_2\mid A)\mathsf P(A) + \mathsf P(T_1=T_2\mid A^\complement)\mathsf P(A^\complement)
\\[1ex] & = {\big(\mathsf P(T_1\mid A)\mathsf P(T_2\mid A)+\mathsf P(T_1^\complement\mid A)\mathsf P(T_2^\complement\mid A)\big)\mathsf P(A) + \ldots\\ \dots  \big(\mathsf P(T_1\mid A^\complement)\mathsf P(T_2\mid A^\complement)+\mathsf P(T_1^\complement\mid A^\complement)\mathsf P(T_2^\complement\mid A^\complement)\big)\mathsf P(A^\complement)}
\end{align}$$
Step 3: Substitute and evaluate.

Why is this different from your answer?  You've assumed that when the child goes to either clinic it may or may not have autism -- each time.  However, this neglects that it is the same child; it either has or does not have autism, but whichever it is, this doesn't change when it visits another clinic.
Hence you don't square the probability of being autistic.  $\mathsf P(A_1\cap A_2)=\mathsf P(A_1) \neq \mathsf P(A_1)\mathsf P(A_2)$

Remark:   Of course there's an error in the assumption of independence.   It works for the general population, but not for the same individual.   Because a child's histology does not change, the probability that two of the same tests on the same individual give the same result will be higher than assumption of independence suggests.   Hence why doctors don't subject patients to the exact same test multiple times to reduce false positive rates is because: it does not work.
