The number of balls in the box is 

\(\displaystyle 1 + 2 + 3 + 4 + 5 +6 + 7 + 8 + 9 + 10 + 2 = 57\).

The number of balls with even numbers is 

\(\displaystyle 2+ 4 + 6 + 8 + 10 = 30\).

Therefore, if a ball is drawn at random, the probability that an even-numbered ball will be selected is

\(\displaystyle \frac{30}{57} = \frac{10}{19}\)

Each of the 26 letters is represented by one red ball; in addition, each of the five vowels is represented by two blue balls for a total of \(\displaystyle 5 \times 2 = 10\) blue balls. The total number of balls is

\(\displaystyle 26 + 10 = 36\).

The probability that a random draw will result in a blue ball being selected is 

\(\displaystyle \frac{10}{36} = \frac{5}{18}\).

Each whole number from one to ten will be represented by a red ball, for a total of ten balls; each even number will be represented by a green ball, for a total of five balls; there will also be five unmarked yellow balls. The number of balls in the box will be \(\displaystyle 10 + 5 + 5 = 20\), 5 of which are green, making the probability of a random draw resulting in a green ball

\(\displaystyle \frac{5}{20} = \frac{1}{4}\).

The number of balls in the box is 

\(\displaystyle 1 + 2 + 3 + 4 + 5 +6 + 7 + 8 + 9 + 10 + 1 = 56\);

The number of odd-numbered balls is 

\(\displaystyle 1+3+ 5 + 7 + 9 = 25\).

Therefore, there are \(\displaystyle 56- 25 = 31\) balls that are not marked with an odd number, making the probability that one of these will be drawn \(\displaystyle \frac{31}{56}\).

To find the probability of an event, we will use the following formula:

\(\displaystyle \text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}\)

Now, we will calculate the following:

\(\displaystyle \text{number of ways event can happen} = 4\)

because there are 4 different ways we can draw a 5 from a deck of cards:

• 5 of hearts
• 5 of diamonds
• 5 of spades
• 5 of clubs

Now, we will calculate the following:

\(\displaystyle \text{total number of possible outcomes} = 52\)

because there are 52 different cards we could potentially draw from a deck. 

So, we will substitute.  We get

\(\displaystyle \text{probability of drawing a 5} = \frac{4}{52}\)

\(\displaystyle \text{probability of drawing a 5} = \frac{2}{26}\)

\(\displaystyle \text{probability of drawing a 5} = \frac{1}{13}\)

Therefore, the probability of drawing a 5 from a deck of cards is \(\displaystyle \frac{1}{13}\).

To find the probability of an event, we will use the following formula:

\(\displaystyle \text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}\)

Now, given the event of calling on a boy, we can calculate:

\(\displaystyle \text{number of ways event can happen} = 17\)

because there are 17 boys in the classroom who can be called on.

We can also calculate the following:

\(\displaystyle \text{total number of possible outcomes} = 31\)

because there are 31 total students (14 girls + 17 boys = 31 students) who could potentially be called on.

Now, we can substitute.  We get

\(\displaystyle \text{probability of calling on a boy} = \frac{17}{31}\)

Therefore, the probability of the teacher calling on a boy is \(\displaystyle \frac{17}{31}\).

Wihtout the clubs, the deck comprises 39 cards, 26 of which are red.

The probability that the first card will be red will be \(\displaystyle \frac{26}{39} = \frac{2}{3}\). The probability that the second will then also be red will be \(\displaystyle \frac{25}{38}\). Multiply the probabilities, and result is \(\displaystyle \frac{2}{3} \cdot \frac{25}{38} =\frac{1}{3} \cdot \frac{25}{19} = \frac{25}{57}\)