Probability of spelling IDAHO correctly I'm working with a problem stated as follows:

A sign reads IDAHO. Two letters are removed and put back at random, each
equally likely to be put upside down as in the correct orientation. What is the
probability that the sign still reads IDAHO?

So, we form our events as:

*

*$A = \{$IDAHO is spelt correctly$\}$

*$B = \{$2 "symmetric" characters are chosen$\}$

*$C = \{$1 "symmetric" and 1 "unsymmetric" character is chosen$\}$

*$D = \{$2 "unsymmetric" characters are chosen$\}$
Where symmetric means that no matter of orientation, the character looks just like it used to before.
Notice, $B,C,D$ form a partition of our sample space. Hence, the the total law of probability gives us:
$$P(A) = P(A|B)P(B) + P(A|C)P(C) + P(A|D)P(D) $$
We trivially know that $P(A|B) = 1/2$, since in choosing two symmetric letters.
Furthermore, $P(B) = \frac{\begin{pmatrix}
3 \\2
\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 3/10$
Then, we have $P(A|C) = 2/4$, since in choosing 1 symmetric letter and 1 unsymmetric, we have to get the correct position on the unsymmetric letter, whereas the symmetric letter can be switched in orientation without changing the word. Meaning we have 2 out of a total 4 possible scenarios. 4 comes from the fact that we can choose to place the symmetric letter in 2 positions, and orient it in 2 ways, however, these are equivalent. Then, the latter unsymmetric letter has already been given a position, but can be oriented in 2 distinct ways.
Furthermore, $P(C) = \frac{\begin{pmatrix}
3 \\2 
\end{pmatrix}\begin{pmatrix}
2 \\1 
\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 3/5$
Lastly,  $P(A|D) = 1/6$, since we only have one scenario in which it can become the same word, and the total scenarios are given as we choose a position for the first letter (2), then we choose orientation (2), the latter letter has already been given a position and orientation can be choosen (2).
Also $P(D) = \frac{\begin{pmatrix}
2 \\
2\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 1/10$
Finally, we have $P(A) = 1/2 \cdot 3/10 + 1/2 \cdot 3/5 + 1/6 \cdot 1/10 = 7/15$, which according to the answer sheet is wrong ($5/16$).
Does anyone of you guys see the mistake. I've tried to find it and gone through my argumentation but can't find the mistake.
Thanks.
 A: It looks like your mistake is in the conditional probabilities. $\mathbb P(A \mid B)$ is fine, but the other two contain errors.
Looking at $\mathbb P(A \mid C)$: we need to require that the asymmetric letter not be flipped (probability $1/2$) and that the two letters not be interchanged (probability $1/2$). Assuming these two events are independent, the probability of both occurring is $1/4$, not $2/4$. (The error in your reasoning here is that there are actually 2 favorable cases out of 8, not 4.)
I claim that a similar mistake is present in $\mathbb P(A \mid D)$; there should be 8 cases to consider, not 6.
EDIT: I should add that I'm interpreting the problem in a specific way, which seems to be indicated by some of your comments (such as the triviality of $\mathbb P(A \mid B)$). Specifically, I'm assuming that:

*

*We remove two letters from IDAHO, leaving two blank spaces;

*we shuffle the letters and replace them in the two blanks;

*the process of shuffling may (or may not) flip them upside down, and whether it does is independent of the positions selected and whether the other letter is flipped.

(I was able to recover the answer of $5/16$ with this interpretation.)
A: This answer identifies the errors in the question’s reasoning, mainly by restating the advice in Aaron Montgomery’s answer and Acccumulation’s comment, then uses all of Acccumulation’s advice to recalculate $P(A)$.
$P(A|B)$ is indeed $\frac{1}{2}$, but let us write it as $\frac{4}{8}$ instead, for consistency with the other probabilities and to follow Acccumulation’s advice.
For $P(A|C)$ and $P(A|D)$, the numbers of favourable outcomes are indeed $2$ and $1$ respectively, but the total numbers of outcomes are incorrect. By your own reasoning, they are both $2 \cdot 2 \cdot 2 = 8$.
One small error was not mentioned by the other users: the expression for $P(C)$ should say $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ instead of $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$; it so happens that these have the same value.
\begin{align}
P(A) &= P(A|B)P(B) + P(A|C)P(C) + P(A|D)P(D) \\
&= \frac{4}{8} \cdot \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} + \frac{2}{8} \cdot \frac{\begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} + \frac{1}{8} \cdot \frac{\begin{pmatrix} 2 \\ 2 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix}}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{4 \cdot 3 + 2 \cdot 6 + 1}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{12 + 12 + 1}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{25}{80} \\
&= \frac{5}{16}. \\
\end{align}
