You asked in a comment whether it is possible to generalise this, though it is not quite clear in what way you would wish to do so. It is easy enough to calculate the number of ways and probability of any particular pattern: for example

• to get $4$ cards from one suit and $2$ each from three suits and $0$ from the one remaining suit, there are ${5 \choose 4}^1{5 \choose 2}^3{5 \choose 0}^1 \frac{5!}{1!3!1!} = 100000$ ways

• to get $2$ each from all five suits, there are ${5 \choose 2}^5 \frac{5!}{5!} = 100000$ ways too

and so on for the other patterns, each of which is a partition of the number of cards drawn into up to a number of parts no more than the number of suits, with each part no more than the number of cards per suit.

You can then turn these into probabilities by dividing by ${25 \choose 10}$. As lulu has commented and MrPuffer has calculated, the answer to your particular question is the complement of $2$ cards from each of the five suits, i.e. $1-\frac{100000}{3268760} \approx 0.97$.

Using R, you can find the number of ways and probabilities for each possible pattern (and this can easily be adapted for other examples to help you generalise) with

library(partitions)
suits        <- 5
cardspersuit <- 5
cardsdrawn   <- 10
restrictparts <- restrictedparts(cardsdrawn, suits)
limitparts <- restrictparts[, restrictparts[1,] <= cardspersuit] 
ways <- numeric(ncol(limitparts))
for (n in 1:ncol(limitparts)){
    part      <- limitparts[,n]  
    tablepart <- table(part)
    ways[n]   <- prod(choose(cardspersuit, part)) * 
                 factorial(suits) /  prod(factorial(tablepart))
    names(ways)[n] <- paste(part, collapse="+")
    }                
cbind(ways, prob=ways / choose(suits*cardspersuit, cardsdrawn))

giving

             ways         prob
5+5+0+0+0      10 3.059264e-06
5+4+1+0+0    1500 4.588896e-04
5+3+2+0+0    6000 1.835558e-03
4+4+2+0+0    7500 2.294448e-03
4+3+3+0+0   15000 4.588896e-03
5+3+1+1+0   15000 4.588896e-03
4+4+1+1+0   18750 5.736120e-03
5+2+2+1+0   30000 9.177792e-03
4+3+2+1+0  300000 9.177792e-02
3+3+3+1+0  100000 3.059264e-02
4+2+2+2+0  100000 3.059264e-02
3+3+2+2+0  300000 9.177792e-02
5+2+1+1+1   25000 7.648160e-03
4+3+1+1+1  125000 3.824080e-02
4+2+2+1+1  375000 1.147224e-01
3+3+2+1+1  750000 2.294448e-01
3+2+2+2+1 1000000 3.059264e-01
2+2+2+2+2  100000 3.059264e-02

If you changed the number of cards drawn to $5$ you would instead get

           ways         prob
5+0+0+0+0     5 9.410879e-05
4+1+0+0+0   500 9.410879e-03
3+2+0+0+0  2000 3.764352e-02
3+1+1+0+0  7500 1.411632e-01
2+2+1+0+0 15000 2.823264e-01
2+1+1+1+0 25000 4.705439e-01
1+1+1+1+1  3125 5.881799e-02

and the probability of drawing at least three cards from a single suit would become $\frac{10005}{53130}\approx 0.19$.