Loaded die probability The loaded die has the following probabilities: 
$0.3,0.2,0.1,0.1,0.2,0.1$ for $1,2,3,4,5,6$ respectively. The question asks

What is the probability of rolling at least one $1$ and no $2's$ in $4$ rolls.

I solved this problem by extending the negative binomial distribution so that if we let $A$ be the event rolling at least one 1 and no 2's then
$$P(A) = \binom{3}{3}0.3*0.2^3+\binom{4}{2}0.3^2*0.2^2+\binom{5}{1}0.3^3*0.2+0.3^4$$ 
I'm not sure if this is the right answer or the right approach so any help or feedback is much appreciated. 
 A: To get exactly one $1$ and no $2$s, we choose which roll we get the $1$ on ($4$ possibilities), roll that one $1$ (with probability $0.3$), then roll a $3, 4, 5,$ or $6$ on the other three rolls (each with probability $0.5$).  The probability is
$$4(0.3)(0.5)^3.$$
To get exactly two $1$s and no $2$s, the probability is
$$6(0.3)^2(0.5)^2.$$
To get exactly three $1$s and no $2$s, the probability is
$$4(0.3)^3(0.5).$$
To get exactly four $1$s, the probability is
$$(0.3)^4.$$
Adding these up gives $P = 0.3471$.
Another way of writing this (in a form similar to what you have) is
$$P(A) = {4 \choose 1}(0.3)^1(0.5)^3 + {4 \choose 2}(0.3)^2(0.5)^2 + {4 \choose 3}(0.3)^3(0.5)^1 + {4 \choose 4}(0.3)^4(0.5)^0 = 0.3471.$$
Yours is $0.0591$ which isn't the same.  Your first two terms kind of have the right idea, except the other probability should be $0.5$, not $0.2$.  You're looking to not get another $2$, and what you have gives the probability that you do.  The third term doesn't make sense with the $_5C_1$.  The fourth term is correct.
Hope this helps.
A: Well, the probability of throwing no one's and no two's is 
$$p_0=\left(\frac 12\right)^4=\frac {1}{16}=.0625$$
while the probability of throwing no two's is $$p_1=\left(\frac 45\right)^4=\frac {256}{625}=.4096$$
The difference is the probability you seek, hence $$.4096-.0625=.3471$$
