This is not too difficult to calculate by hand. In general, if you are counting an objective value of t on a side die, with the die roll represented as random variable D, the probability of obtaining a specific value is, of course, p (D = t) = 1 / m . If you are trying to get t or more on an individual roll, you have probability of success of p (D> = t) = (m-t + 1) / m. For example, its probability of 7 or more in a d10 is P (D> = 7) = (10-7 + 1) / 10 = 4/10. In AnyDice this would be 1d10> = 7.
By throwing multiple dice rolls independently (and not adding them), we can calculate the probabilities of a particular combination by multiplication. For example, the probability of getting two rolls higher than 7 and then two rolls lower than 7, in that order, is P (D1> = 7) * P (D2> = 7) * P (D3 <7) * P (D3 < 7) = 0.4*0.4*0.6*0.6. On AnyDice this would be 1d10>= 7 and 1d10> = 7 and 1d10 <7 and 1d10 <7.
However, we do not want dice rolls in a specific, ordered combination, such as two successes and two failures. Any combination of two successes and two failures will serve. Therefore, we need the number of ways to obtain exactly two successes in four runs, which use the rule of "number of combinations" or nCr. 4 choose 2 = 6, so multiply our previous 0.4 * 0.4 * 0.6 * 0.6 by 6 to get the odds of getting at least 2 rolls of at least 7 in 4d10 is 34.56%.
The general formula here is a binomial distribution:
p_success ^ num_successes * p_failure ^ num_failures * nCr (num_failures + num_successes, num_successes)
In AnyDice, the easiest way I know to get combinations like this is to compare first and then add the variables. Therefore, if you write a formula like this, you will get the distribution for the count of 4d10 rolls at least 7.
output (1d10> = 7) + (1d10> = 7) + (1d10> = 7) + (1d10> = 7)
4d output (1d10> = 7)
For a large number of dice rolls (at least 14 for our example d10> = 7), the normal distribution is a good approximation of these results. Note that the related distribution of the total of several dice rolls is given in Wolfram and Stats SE, or they are fairly trivial to calculate in Anydice with formulas like (3d6 + 1d8)> 10.