A set of Domino blocks consists of $28$ blocks 
A set of Domino blocks consists of $28$ blocks, each block corresponds to an unordered pair of numbers $(i, j)$, where $i, j = 0,1,2,3,4,5,6$.
We say that we can add a block $B(k,l)$ to block $A(i,j)$ if $k=i$ or $k=j$ or $l=i$ or $l=j$. We arrange the two matching blocks so that the identical numbers are next to each other, for example: $A(1,2)B(2,0)$. The next block can be added to the received string, if there is a number on it equal to one of the extremes of the received string (in example number $1$ or $0$). We draw three blocks $K$, $L$, $M$ without returning. Calculate the probability, that we can add $M$ to the string made up of $K$ and $L$ blocks if it is known that $K$ and $L$ bricks fit together.


This is what I know so far:
Let $A_1$ be the probability that K or L (either one of these, both cannot be) are from the kind of set of block which have the two same numbers on each side (like (0,0), (1,1), ...) and that M fits to either K or L
Let $A_2$ be the probability that K and L are not from this kind of set I've talked about (this means K and L are not from {(0,0), (1,1), ...}) and that M fits to either K or L
The probability for both cases is $\frac{11}{26}$, just take an example and you'll see that that's the case for M, in both cases we've got 11 possibilities for M within 26 possible blocks.
Let $B$ be the probability that M fits either K or L
$P\left(M\:fits\right)=P\left(M\:fits\:|\:L\:or\:K\:are\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(L\:or\:K\:are\:from\:\left\{\left(0,0\right),...\right\}\right)+P\left(M\:fits\:|\:L\:and\:K\:are\:not\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(L\:and\:K\:are\:from\:not\:from\left\{\left(0,0\right),...\right\}\right)=P\left(M\:fits\:\cap \:L\:or\:K\:are\:from\:\left\{\left(0,0\right),...\right\}\right)\:+\:P\left(M\:fits\:\cap \:\:L\:and\:K\:are\:not\:from\:\left\{\left(0,0\right),...\right\}\right)\:$
Which essentially means
$P\left(B\right)=P\left(A_1\right)+P\left(A_2\right)=\frac{22}{26}$
But this is not true! It should be $\frac{11}{26}$
What am I doing wrong?
This is the solution I've found:
$$P\left(M\:fits\right)=P\left(M\:fits\:\cap \:K\:and\:L\:are\:not\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(K\:and\:L\:are\:not\:from\:\left\{\left(0,0\right),...\right\}|M\:fits\right)+P\left(M\:fits\:\cap \:\:K\:or\:L\:are\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(K\:or\:L\:are\:\:from\:\left\{\left(0,0\right),...\right\}|M\:fits\right)$$
But I don't see why this should be the solution in the first place, from where is this formula coming from?
 A: I think the solution is wrong. It should be:
$$P\left(M\:fits|K\:and\:L\:fit\right)=P\left(M\:fits\:\cap \:K\:and\:L\:are\:not\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(K\:and\:L\:are\:not\:from\:\left\{\left(0,0\right),...\right\}|K\:and\:L\:fit\right)+P\left(M\:fits\:\cap \:\:K\:or\:L\:are\:from\:\left\{\left(0,0\right),...\right\}\right)P\left(K\:or\:L\:are\:\:from\:\left\{\left(0,0\right),...\right\}|K\:and\:L\:fit\right)$$
Basically, you have already said that, if K and L fit, then (in any case) the probability that M will fit is 11/26.
You should not ADD those together.
Here's another problem to help you understand why:
Imagine that you flip a penny, a nickel, and a dime, and you want to compute the probability that they are all the same (all heads or all tails) given that the penny and the nickel ARE the same.
If you reasoned the way that you did above, you would argue that
either the penny and nickel are both heads, and the probability that the dime is too is 1/2;
or the penny and nickel are both tails, and the probability that the dime is too is 1/2;
So the probability that they are all the same is 1/2 + 1/2 = 1 (which is not true)
If you want to compute the probability $P(A|B)$, and the event $B$ can be decomposed into two mutually exclusive events $B_1 \cup B_2 = B$, then
$$P(A|B) = P(A|B_1)P(B_1|B) + P(A|B_2)P(B_2|B)$$
This is the law of total probability.
You can consider this formula to be the same as:
$$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2)$$
if you take event all events to include the given information.
For your example:
Let A be the event that M fits given K and L fit (what you're trying to find)
Let $B_1$ be the event that $K\:and\:L\:are\:not\:from\:\left\{\left(0,0\right),...\right\}\:given\: K\:and\:L\: fit$.
Let $B_2$ be the event that $K\:and\:L\:are\:from\:\left\{\left(0,0\right),...\right\}\:given\: K\:and\:L\: fit$.
In general, to use this formula, you would normally need to compute the probabilities $P(B_1)$ and $P(B_2)$. However, in this case they are both multiplied by 11/26. Since you know that $P(B_1)$ and $P(B_2)$ add to one (exactly one of the two events has to occur, given that K and L fit), then
$$(11/26)P(B_1) + (11/26)P(B_2) = 11/26$$
