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 $8$
 A: 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:
$1-\sum\limits_{k=0}^{2}\dfrac{\binom{n}{k}\cdot3^{n-k}}{4^n}=0.3214$
A: The answer 8 given is  the correct answer (assuming you want the probability of at least 3 twos to be at least .3214).
EDIT I had misread the required probability, thinking for some reason that it was .3294.  The answer 8 is in fact correct.
Here is a good way to attack the problem:  The probability Q(n) of at most 2 twos in $n$ rolls is given by the probability of no twos, plus probability of 1 two, plus probability of 2 twos, and that needs to be less than or equal to .6786.  By the binomial distribution, these three numbers are:
$$\left(\frac{3}{4}\right)^n + n \left(\frac{3}{4}\right)^{n-1}\frac14
+ \frac{n(n-1)}{2} \left(\frac{3}{4}\right)^{n-2} \left(\frac{`}{4}\right)^2
$$
A bit of algebra cleans this up to 
$$
Q(n) = \left(\frac{3}{4}\right)^n \frac{18+5n+n^2}{18} \leq .6786
$$
And if you tabulate that you get 
$$
Q(6) = .8306\\
Q(7) = .7564 \\
Q(8) = .6785  < .6786 \\
Q(9) = .6007
$$
If you do the problem using the normal distribution approximation to the binomial distribution, then it looks like you get 8 as an answer.
