Finding the probability of a selecting at least 1 of an element. If there are 18 red and 2 blue marbles what is the probability of selecting 10 marbles where there is either 1 or 2 blue marbles in selected set.
Also, it seems intuitive that the probability should be twice that of selecting a single blue marble in a set of 10 from 19 red and 1 blue, but I'm not sure.

Given $18$ red marbles and $2$ blue marbles:


*

*The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$

*The number of ways to choose $10$ red marbles is $\binom{18}{10}=43758$

*Hence the probability to choose only red marbles is $\frac{43758}{184756}=\frac{9}{38}$

*Hence the probability to choose not only red marbles is $1-\frac{9}{38}\approx76.31\%$

Given $19$ red marbles and $1$ blue marble:


*

*The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$

*The number of ways to choose $10$ red marbles is $\binom{19}{10}=92378$

*Hence the probability to choose only red marbles is $\frac{92378}{184756}=\frac{1}{2}$

*Hence the probability to choose not only red marbles is $1-\frac{1}{2}=50\%$

As you can see, the former is not twice the latter...
