Probability of getting a red ball when picking 5 from 10 in a bag There are 10 balls in the bag, 2 of them are red. We are picking randomly 5 balls from 10, what is the probability that at least one of picken 5 is red?
So far I found out that are $^{10}C_5 = 252 $ possibilties of picking 5 from 10.
It says to find the probability of picking at least $1$ red ball. So your total probability is simply, 
$$P(X\geq1)=P(X=1) + P(X=2)$$ 
Now, this is a simply fancy notation for saying it is the sum of the probability that you pick $1$ red ball out of the $5$ and $2$ red balls out of the $5$. Then for $1$ red ball we get, $$P(X=1) = \frac{^2C_1}{^{10}C_5} \;\text{or} \;\frac{\binom21}{\binom {10}{5}}$$ 
And for $2$ red balls we get,
 $$P(X=2) =  \frac{^2C_2}{^{10}C_5}\;\text{or} \;\frac{\binom22}{\binom {10}{5}}$$ The sum of these two gives you your answer.
It says to find the probability of picking at least $1$ red ball. So your total probability is simply, 
$$P(X\geq1)=P(X=1) + P(X=2)$$ 
Now, this is a simply fancy notation for saying it is the sum of the probability that you pick $1$ red ball out of the $5$ and $2$ red balls out of the $5$. Then for $1$ red ball we get, $$P(X=1) = \frac{^2C_1}{^{10}C_5} \;\text{or} \;\frac{\binom21}{\binom {10}{5}}$$ 
And for $2$ red balls we get,
 $$P(X=2) =  \frac{^2C_2}{^{10}C_5}\;\text{or} \;\frac{\binom22}{\binom {10}{5}}$$ The sum of these two gives you your answer.