Aregular 6-sided die is tossed 5 times what is the probability that in two of those tosses we get a number less than 3

In probability and statistics, there is an equation used for repeated trials. In this equation, you find the probability of getting 'r' successful events out of 'n' trials. Moreover, you should incorporate the probability of the success per trial. For a 6-sided face, each side has a probability of success of 1/6. On the other hand, the probability of each side not appearing is 5/6, because when you add these two, it would sum up to 1 as always.

Now, the equation for repeated trials is:

Total Probability = n!/r!(n-r)! * p^(n-r) * q^r

where
n = 5 tosses
r = 1 and 2 (since you want to see the probability of a side less than 3)
p = 1/6
q = 5/6

So, you add the individual probability when r=1, and when r=2.

Total Probability = 5!/1!(5-1)! * (1/6)^(5-1) * (5/6)^1 + 5!/2!(5-2)! * (1/6)^(5-2) * (5/6)^2
Total Probability = 50/243 or 20.58%

I'm not sure exactly but my best guess would be 1/4