Picking two numbers - Calculate the probability We take randomly two numbers without replcement from a box that contains the numbers $1,2,\ldots , 10$.
Calculate the probability of the following events :
a) A = their sum is $11$
b) B = their product is even
c) C = the smaller one is $4$ or $5$
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I have done the following :
a) To get the sum $11$ we can have the $5$ combinations :  $1+10$ , $2+9$ , $3+8$ , $4+7$ , $5+6$
In total there are $10\cdot 9=90$ possible combinations.
So we have the probability $P(A)=\frac{5}{90}=\frac{1}{18}$.
b) The product is even if at least one of the two numbers is even.
So for the first number there are $10$ possible number to take.
In total there are $10\cdot 9=90$ possible combinations.
So we have the probability $P(B)=\frac{5}{90}=\frac{1}{18}$.
c) The smaller number is $4$ or $5$. So do we have to calculate the probability of the union of $E_1$ : the smaller one is $4$ and $E_2$ : the smaller one is $5$ ?

For $a)$, please note that we could choose any of the $10$ numbers as the first number. Then there is exactly $1$ second number out of remaining $9$ that gives sum of $11$.
So the probability is $\frac{1}{9}$.
In the way you wrote, you should have also counted $10 + 1, 9 + 1$ etc. and the answer should have been $\frac{10}{90}$
For $b)$, only if you pick both numbers odd, you do not get an even number as product. So the probability is,
$1 - \frac{5}{10} \cdot \frac{4}{9}$
$c)$ yes, you have to simply add probabilities for both cases - when the smaller number is $4$ and when the smaller number is $5$.
