<s>[INST] Solve the following problem by using the Binomial formula If $n=10$ and $p=0.60$ find $P(X\geq 3)$. 
The formula we've been given: $P(X=x) = \binom{n}{x} \cdot  p^x \cdot  q^{n-x}$, where $n$ is the number of trials, $p$ is the probability of success and $q = 1-p$.
I know that I could do $P(X\geq 3) = \binom{10}3 \cdot  0.6^3 \cdot  0.4^7$ adding until I reach $\binom{10}{3} \cdot  0.60^{10} \cdot  0.40^0$.
I know there is a way I could do this without having to individually calculate from 3 until 10. I'm thinking along the lines of 
$P(X<3) = \binom{10}{2}  \cdot  0.4^2 \cdot  0.6^8 + \binom{10}{1}  \cdot  0.4^1 \cdot  0.6^9 + \binom{10}{0}  \cdot  0.4^0 \cdot  0.6^{10}$ 
But I'm having doubts about this because I tried the long way and got $0.98771$ but with the other method I get $0.1673$.
 [/INST] You have exchanged $p$ and $1-p$. As you said $p=0.6$. Therefore 
$P(X\leq 2)={10 \choose 2} \cdot  0.6^2 \cdot 0.4^8 + {10 \choose 1} \cdot 0.6^1 \cdot 0.4^9 + {10 \choose 2} \cdot 0.6^0 \cdot 0.4^{10}=0.01229$
Thus $P(X\geq 3)=1-P(X\leq 2)=1-0.01229=0.98771$
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