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.
 A: Another approachment is through hypergeometric distribution.


*

*The probability that there is exactly one blue ball in the selected set is:
$$P(A_1)=\dfrac{\color{blue}{\dbinom 2 1}\cdot \color{red}{\dbinom {18} {10-1}}}{\dbinom {20}{ 10} }$$

*The probability that there are exactly  2 blue balls in the selected set is:
$$P(A_2)=\dfrac{\color{blue}{\dbinom 2 2}\cdot \color{red}{\dbinom {18} {10-2}}}{\dbinom {20}{ 10} }$$
So, the probability you are looking for is $P(A_1)+P(A_2)\approx 0.763 $
A: In this scenario, with no restrictions, there are 3 possible cases:
$\bullet$ When selecting 10, there were no blue marbles ($A_{none}$)
$\bullet$ When selecting 10, there was one blue marble ($A_1$)
$\bullet$ When selecting 10, there were two blue marbles ($A_2$)
As there are no other possibilities, it stands to reason then that these three events form a partition of our overall sample space.  That is: $X = A_{none}\cup A_1 \cup A_2$ and each $A$ is pairwise disjoint from the others.
It follows then, that $1 = P(X) = P(A_{none}) + P(A_1) + P(A_2)$ by rule of sums.
The event you are interested in is that either one or two blue marbles were selected, namely $A_1\cup A_2$.
By the above, $P(A_1\cup A_2) = 1 - P(A_{none})$
To calculate $P(A_{none})$, for each of the 10 marbles selected, you will have selected red.  So, multiply the probabilities for each step that you in fact picked red.
So: $P(A_{none}) = \dfrac{18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11} = \dfrac{10\cdot 9}{20\cdot 19}$
And finally $P(A_1\cup A_2) = 1 - P(A_{none}) = 1 - \frac{10\cdot9}{20\cdot 19}\approx 0.763$
In comparison, if there were 19 red and 1 blue, it winds up being $1 - \frac{10}{20} = 0.5$  The answer to the original question is in fact a bit more than 50% more likely.
A: 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...
