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Let's work through the problems one by one. Here are the interpretations and solutions:

### Problem 1:
**A bag contains 7 red balls and 5 white balls. 4 balls are drawn at random. Find the probability that:**
1. **All of them are red.**
2. **Two of them are red and two are white.**

#### Solution:
- Total number of balls = 7 (red) + 5 (white) = 12
- Total ways to draw 4 balls from 12: \( \binom{12}{4} \)

**(i) Probability that all 4 balls are red:**
- Ways to draw 4 red balls from 7: \( \binom{7}{4} \)
- Probability = \( \frac{\binom{7}{4}}{\binom{12}{4}} \)

**(ii) Probability that 2 balls are red and 2 are white:**
- Ways to draw 2 red balls from 7: \( \binom{7}{2} \)
- Ways to draw 2 white balls from 5: \( \binom{5}{2} \)
- Probability = \( \frac{\binom{7}{2} \cdot \binom{5}{2}}{\binom{12}{4}} \)

### Problem 2:
**A pair of fair dice is thrown. Find the probability that the sum is 4 or less, given that the two numbers appearing are different.**

#### Solution:
- Total outcomes when rolling two dice: \( 6 \times 6 = 36 \).
- To find the probability that the sum is 4 or less, let's list possible pairs (1, 2), (2, 1), (1, 3), (3, 1), (2, 2).
- From these, exclude (2, 2) since we want different numbers.
- Count remaining pairs with sum 4 or less and divide by the total outcomes with different numbers.

### Problem 3:
**If \( P(A \cup B) = \frac{5}{6} \), \( P(A \cap B) = \frac{1}{3} \), and \( P(A') = \frac{1}{2} \), find \( P(A) \) and \( P(B) \). Hence, show that A and B are independent.**

#### Solution:
- Use the formula for the complement: \( P(A') = 1 - P(A) \).
- Calculate \( P(A) \) and \( P(B) \) using the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
- A and B are independent if \( P(A \cap B) = P(A) \times P(B) \). Verify this condition.

### Problem 4:
**A and B are two weak students of statistics, and their chances of solving a problem correctly are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If the probability of them making a common error is \(\frac{1}{25}\), find the probability they obtain the same answer.**

#### Solution:
- Calculate the probability that both solve it correctly or both solve it incorrectly.
- Use the probability of a common error to help find the relevant probability for incorrect answers.

Would you like more detailed solutions or explanations for any of these problems? Here are some related questions to deepen your understanding:

1. How would you calculate probabilities if the conditions of drawing were changed to "with replacement"?
2. What is the significance of events being independent in probability?
3. How does the concept of conditional probability apply to rolling dice?
4. What is the importance of understanding combinations (\( \binom{n}{r} \)) in probability calculations?
5. How can we interpret the probability of a "common error" in the context of problem-solving?

**Tip:** When dealing with multiple events, always break down the problem into smaller parts to ensure you’re accounting for all outcomes.

Solve all the problems.
1. A bag contains 7 red balls and 5 white balls. 4 balls are drawn at random. What is the probability that: (i) all of them are red, (ii) two of them would be red and two white?
2. A pair of fair dice is thrown. Find the probability that the sum is 4 or less, given that the two numbers appearing are different.
3. If P(A ∪ B) = 5/6, P(A ∩ B) = 1/3 and P(A') = 1/2, find P(A) and P(B). Hence, show that A and B are independent.
4. A and B are two weak students of statistics and their chances of solving a problem correctly are 1/2 and 1/3, respectively. If the probability of them making a common error is 1/25, find the probability they obtain the same answer.

Solve problems on probability involving drawing balls from a bag, dice rolls, and the application of set operations and event independence. This exercise is ideal for undergraduate students, focusing on combinations and probability theory.

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