die roll probability with four sided die How many times must you roll a four sided die so that the probability of getting at least $3$ twos is $0.3214$.
$$P(n \ge 3)=0.3214$$
$$P(x)=(1/2)^3=1/8$$
answer given is $8$

The probability of getting at least $3$ twos out of $n$ rolls is $\sum\limits_{k=3}^{n}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}$
Using trial & error, we get $\sum\limits_{k=3}^{n}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}=0.3214 \implies n=8$

You can also calculate $1$ minus the probability of the complementary event:
$1-\sum\limits_{k=0}^{2}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}=0.3214$
