Answer:
If all you care about is whether you roll 2 or not, you get a Binomial distribution with an individual success probability 1/6. The probability of rolling 2 at least two times, is the same as the probability of not rolling 2 at zero or one time.
the answer is, 1 - bin(k=0, n=4, r=1/6) - bin(k=1, n=4, r=1/6). This evaluates to about 13%, just like your result (you just computed all three outcomes satisfying the proposition rather than the two that didn’t).
Step-by-step explanation:
