From the set S = 3, 5, 6, 8, 11, 14, 17, find the probability that a randomly selected number is even, prime, greater than 10, or a multiple of 3.

The set is $S = \{3, 5, 6, 8, 11, 14, 17\}$.
The total number of elements in the set is $n(S) = 7$.

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

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

The probability that the number is even is $\boxed{\frac{3}{7}}$.

(ii) Probability that the number is prime.
The prime numbers in $S$ are numbers greater than 1 with only two divisors: 1 and themselves.
The prime numbers in $S$ are $\{3, 5, 11, 17\}$.
The number of prime numbers is $n(\text{prime}) = 4$.
The probability is:

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

The probability that the number is prime is $\boxed{\frac{4}{7}}$.

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

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

The probability that the number is greater than 10 is $\boxed{\frac{3}{7}}$.

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

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

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

A bag contains:

• 4 blue balls (B)
• 3 green balls (G)
• 3 yellow balls (Y) Total number of balls = $4 + 3 + 3 = 10$. One ball is drawn and not replaced, then a second ball is drawn.

(i) Probability of obtaining two yellow balls.
Probability of drawing a yellow ball first ($Y_1$):

$$P(Y_1) = \frac{\text{Number of yellow balls}}{\text{Total balls}} = \frac{3}{10}$$

After drawing one yellow ball, there are 2 yellow balls left and 9 total balls.
Probability of drawing a second yellow ball ($Y_2$) given the first was yellow:

$$P(Y_2 | Y_1) = \frac{\text{Number of yellow balls remaining}}{\text{Total balls remaining}} = \frac{2}{9}$$

Probability of two yellow balls:

$$P(Y_1 \text{ and } Y_2) = P(Y_1) \times P(Y_2 | Y_1) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$

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

(ii) Probability of obtaining two blue balls or two green balls.
This is $P(\text{two blue balls}) + P(\text{two green balls})$.
Probability of two blue balls ($B_1$ and $B_2$):

$$P(B_1) = \frac{4}{10}$$ $$P(B_2 | B_1) = \frac{3}{9}$$ $$P(B_1 \text{ and } B_2) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$

Probability of two green balls ($G_1$ and $G_2$):

$$P(G_1) = \frac{3}{10}$$ $$P(G_2 | G_1) = \frac{2}{9}$$ $$P(G_1 \text{ and } G_2) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$

Probability of two blue balls or two green balls:

$$P(\text{two blue or two green}) = P(B_1 \text{ and } B_2) + P(G_1 \text{ and } G_2) = \frac{2}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5}$$

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

(iii) Probability of obtaining one blue ball and one green ball in any order.
This means $P(B_1 \text{ and } G_2) + P(G_1 \text{ and } B_2)$.
Probability of drawing a blue ball first, then a green ball ($B_1$ and $G_2$):

$$P(B_1) = \frac{4}{10}$$

After drawing one blue ball, there are 3 green balls left and 9 total balls.

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

Probability of drawing a green ball first, then a blue ball ($G_1$ and $B_2$):

$$P(G_1) = \frac{3}{10}$$

After drawing one green ball, there are 4 blue balls left and 9 total balls.

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

Probability of one blue ball and one green ball in any order:

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

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