A set of numbers is S = 3, 5, 6, 8, 11, 14, 17. Find the probability of choosing an even number, a prime number, a number greater than 10, and a multiple of 3.

The even numbers in $S$ are $\{6, 8, 14\}$.
The number of even numbers is $n(\text{even}) = 3$.
The probability of choosing an even number is:

$$P(\text{even}) = \frac{n(\text{even})}{n(S)} = \frac{3}{7}$$

The probability of choosing an even number is $\boxed{\frac{3}{7}}$.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
The prime numbers in $S$ are $\{3, 5, 11, 17\}$.
The number of prime numbers is $n(\text{prime}) = 4$.
The probability of choosing a prime number is:

$$P(\text{prime}) = \frac{n(\text{prime})}{n(S)} = \frac{4}{7}$$

The probability of choosing a prime number is $\boxed{\frac{4}{7}}$.

The numbers in $S$ greater than 10 are $\{11, 14, 17\}$.
The number of elements greater than 10 is $n(>10) = 3$.
The probability of choosing a number greater than 10 is:

$$P(>10) = \frac{n(>10)}{n(S)} = \frac{3}{7}$$

The probability of choosing a number greater than 10 is $\boxed{\frac{3}{7}}$.

The multiples of 3 in $S$ are $\{3, 6\}$.
The number of multiples of 3 is $n(\text{multiple of 3}) = 2$.
The probability of choosing a multiple of 3 is:

$$P(\text{multiple of 3}) = \frac{n(\text{multiple of 3})}{n(S)} = \frac{2}{7}$$

The probability of choosing a multiple of 3 is $\boxed{\frac{2}{7}}$.

Part 11 (b):
A bag contains 4 blue balls, 3 green balls, and 3 yellow balls.
The total number of balls is $4 + 3 + 3 = 10$.
One ball is drawn at random and not replaced. A second ball is then drawn.

The probability of drawing a yellow ball first is $\frac{3}{10}$.
After drawing one yellow ball, there are 2 yellow balls left and 9 total balls left.
The probability of drawing a second yellow ball is $\frac{2}{9}$.
The probability of drawing two yellow balls is:

$$P(\text{two yellow balls}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$

The probability of obtaining two yellow balls is $\boxed{\frac{1}{15}}$.

The probability of drawing two blue balls is:

$$P(\text{two blue balls}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$

The probability of drawing two green balls is:

$$P(\text{two green balls}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$

The probability of drawing two blue balls or two green balls is the sum of their individual probabilities:

$$P(\text{two blue or two green balls}) = P(\text{two blue balls}) + P(\text{two green balls})$$ $$P(\text{two blue or two green balls}) = \frac{2}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5}$$

The probability of obtaining two blue balls or two green balls is $\boxed{\frac{1}{5}}$.

This means either (blue first, then green) or (green first, then blue).
The probability of drawing a blue ball first, then a green ball:

$$P(B_1 \text{ and } G_2) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$

The probability of drawing a green ball first, then a blue ball:

$$P(G_1 \text{ and } B_2) = \frac{3}{10} \times \frac{4}{9} = \frac{12}{90} = \frac{2}{15}$$

The probability of obtaining one blue ball and one green ball in any order is the sum of these probabilities:

$$P(\text{one blue and one green}) = P(B_1 \text{ and } G_2) + P(G_1 \text{ and } B_2)$$ $$P(\text{one blue and one green}) = \frac{2}{15} + \frac{2}{15} = \frac{4}{15}$$

The probability of obtaining one blue ball and one green ball in any order is $\boxed{\frac{4}{15}}$.