Worksheet on probability problems involving dice, cards, and spinners.

It seems you've uploaded an image containing a series of probability problems. Since the image itself isn't directly visible in this context, I'll infer that the problems are related to probability and combinatorics based on your description. Below, I will solve each problem step by step, assuming typical scenarios for such questions.

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Problem 1:

Ranger is part of a class of 30 students who have been given the task...

#### Solution:
This problem is incomplete as it doesn't specify the task or the details of what Ranger needs to do. However, if we assume a common scenario (e.g., Ranger is chosen randomly from the class), the probability of Ranger being selected can be calculated as follows:

- Total number of students = 30
- Probability of Ranger being selected = \( \frac{1}{30} \)

If more details are provided about the task, please clarify so I can refine the solution.

---

Problem 2:

Drawing two cards from a deck of standard cards...

#### Solution:
Assuming the task is to draw two cards from a standard deck of 52 cards without replacement, let's calculate the probability of a specific event (e.g., drawing two aces).

1. Total number of ways to draw 2 cards from 52:
\[
\binom{52}{2} = \frac{52 \times 51}{2} = 1326
\]

2. Number of ways to draw 2 aces from the 4 aces in the deck:
\[
\binom{4}{2} = \frac{4 \times 3}{2} = 6
\]

3. Probability of drawing 2 aces:
\[
P(\text{2 aces}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{1326} = \frac{1}{221}
\]

If the task involves a different condition (e.g., drawing one ace and one king), please specify.

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Problem 3:

A small deck of cards allows a drawer to choose any card at random...

#### Solution:
This problem is incomplete as it doesn't specify the total number of cards in the deck or the condition for success. Assuming a standard deck of 52 cards and asking for the probability of drawing a specific type of card (e.g., a heart):

- Total number of cards = 52
- Number of hearts = 13
- Probability of drawing a heart:
\[
P(\text{Heart}) = \frac{13}{52} = \frac{1}{4}
\]

If the deck size or condition differs, please provide more details.

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Problem 4:

Two coins tossed consecutively four times...

#### Solution:
Assuming the task is to find the probability of getting heads on both tosses in all four trials:

1. Probability of getting heads on one coin toss:
\[
P(\text{Heads on one toss}) = \frac{1}{2}
\]

2. Probability of getting heads on both coins in one trial:
\[
P(\text{Heads on both coins}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
\]

3. Probability of getting heads on both coins in all four trials:
\[
P(\text{Heads on both coins in all 4 trials}) = \left( \frac{1}{4} \right)^4 = \frac{1}{256}
\]

If the task involves a different condition (e.g., at least one heads), please specify.

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Problem 5:

In a certain town, Peter has 8 coins when he leaves home...

#### Solution:
This problem is incomplete as it doesn't specify the condition or the nature of the coins (e.g., fair coins, biased coins). Assuming Peter flips all 8 coins and we need to find the probability of getting exactly 5 heads:

1. Total number of outcomes when flipping 8 coins:
\[
2^8 = 256
\]

2. Number of ways to get exactly 5 heads:
\[
\binom{8}{5} = \frac{8!}{5! \cdot 3!} = 56
\]

3. Probability of getting exactly 5 heads:
\[
P(\text{Exactly 5 heads}) = \frac{\binom{8}{5}}{2^8} = \frac{56}{256} = \frac{7}{32}
\]

If the task involves a different condition, please clarify.

---

Problem 6:

What is the probability of a full court when a point is served?

#### Solution:
This problem is incomplete as it doesn't specify the rules or conditions of the game (e.g., tennis, volleyball). Assuming a standard tennis match where a "full court" means the ball lands within the boundaries of the court:

- If the server always serves within the court boundaries, the probability is 1 (or 100%).
- If there is a chance of serving out of bounds, additional information is needed (e.g., the server's accuracy rate).

Please provide more details.

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Problem 7:

A pack of cards is shuffled... What is the probability that the top card is a club?

#### Solution:
Assuming a standard deck of 52 cards:

- Total number of cards = 52
- Number of clubs = 13
- Probability of the top card being a club:
\[
P(\text{Top card is a club}) = \frac{13}{52} = \frac{1}{4}
\]

---

Problem 8:

What is the probability that the sum of the numbers rolled on two dice is divisible by 3?

#### Solution:
When rolling two six-sided dice, the possible sums range from 2 to 12. We need to count the sums that are divisible by 3 (i.e., 3, 6, 9, 12).

1. Possible outcomes for each sum:
- Sum = 3: (1, 2), (2, 1) → 2 outcomes
- Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → 5 outcomes
- Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes
- Sum = 12: (6, 6) → 1 outcome

2. Total number of favorable outcomes:
\[
2 + 5 + 4 + 1 = 12
\]

3. Total number of possible outcomes:
\[
6 \times 6 = 36
\]

4. Probability of the sum being divisible by 3:
\[
P(\text{Sum divisible by 3}) = \frac{12}{36} = \frac{1}{3}
\]

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Final Answer:

If you can provide more details or clarify any of the problems, I can refine the solutions further. Based on the assumptions made above, the answers are:

1. \( \boxed{\frac{1}{30}} \) (if selecting Ranger randomly)
2. \( \boxed{\frac{1}{221}} \) (probability of drawing 2 aces)
3. \( \boxed{\frac{1}{4}} \) (probability of drawing a heart)
4. \( \boxed{\frac{1}{256}} \) (probability of getting heads on both coins in all 4 trials)
5. \( \boxed{\frac{7}{32}} \) (probability of getting exactly 5 heads in 8 coin flips)
6. Needs clarification.
7. \( \boxed{\frac{1}{4}} \) (probability of the top card being a club)
8. \( \boxed{\frac{1}{3}} \) (probability of the sum being divisible by 3)

If you have specific details or conditions for any problem, feel free to share them!