There are 3 red balls and 7 blue balls in a bag. We take 4 balls randomly from the bag without replacement. What is the probability of getting at least $2$ blue balls?
When it asks for specific quantities of balls, i.e. $2$ red balls and $2$ blue balls, I understand to do 
$\dfrac{\binom{3}{2}\cdot\binom{7}{2}}{\binom{10}{4}}$
But how do I solve this when it says at least? I have tried computing $\binom{7}{2} + \binom{7}{3} + \binom{7}{4} \,etc. / \binom{10}{4}$, but this is incorrect.
The correct answer is $\frac{29}{30}$. Could someone please provide the steps on how to get to this solution?
 A: Let $X$ be the random variable denoting how many blue balls are selected.
The question asks us to find $Pr(X\geq 2)$
Breaking apart via cases:
$Pr(X\geq 2)=Pr(X=2)+Pr(X=3)+Pr(X=4)$
In other words, to have at least two, this is the same as asking having exactly two or exactly three or exactly four.
You say you know how to calculate each of these specific cases.
It continues then as:
$$Pr(X\geq 2)=\frac{\binom{3}{2}\binom{7}{2}+\binom{3}{1}\binom{7}{3}+\binom{3}{0}\binom{7}{4}}{\binom{10}{4}}$$
Each term of which comes from multiplication principle:  Pick the red balls followed by pick the blue balls.

Alternatively, one can see that $Pr(X\geq 2) = 1-Pr(X<2) = 1-Pr(X=1)$ in this specific case since $Pr(X=0)=0$ (there are only three red balls to pick, we can't have picked four red)
We have then $1-\frac{\binom{3}{3}\binom{7}{1}}{\binom{10}{4}} = 1-\frac{1}{30}=\frac{29}{30}$
A: The probability of getting  at least two blue balls is equal to $$1-\mathbb{P}(\text{getting at most one 1 blue ball})=1-\mathbb{P}(\text{getting 1 blue ball})-\mathbb{P}(\text{getting 0 blue balls}).$$
