Quantity I- In a question paper probability of a question not being answered by three students is `0,5, 0.4,0.7` respectively. Find the probability that at most two students will solve the question.
Quantity II- Dipak have 5 black balls & 7 white balls, in his bag. If three balls are drawn at random from bag, then find probability of getting at least 1 black ball.
Quantity III- nirmal speaks the turth 4 out of 5 times, and Puja speaks the truth 6 out of 7 times. What is the probability that they will contradict each othre in stating the same fact?

Options

A. (@,&)
B. (#,@)
C. (@,@)
D. (**,#)

Explanation

To solve the problem, we will analyze each quantity step by step.

### Quantity I:
We need to find the probability that at most two students will solve a question given the probabilities that each student does not answer it.

1. **Identify the probabilities of not answering:**
- Student A: 0.5
- Student B: 0.4
- Student C: 0.7

2. **Calculate the probabilities of answering:**
- Probability that Student A answers = 1 - 0.5 = 0.5
- Probability that Student B answers = 1 - 0.4 = 0.6
- Probability that Student C answers = 1 - 0.7 = 0.3

3. **Calculate the probability that all three students answer the question:**
- P(All answer) = P(A answers) * P(B answers) * P(C answers)
- = 0.5 * 0.6 * 0.3 = 0.09

4. **Calculate the probability that at most two students answer:**
- P(At most 2 answer) = 1 - P(All answer)
- = 1 - 0.09 = 0.91

### Quantity II:
We need to find the probability of getting at least one black ball when drawing three balls from a bag containing 5 black balls and 7 white balls.

1. **Total balls in the bag:**
- Total = 5 (black) + 7 (white) = 12 balls

2. **Calculate the probability of drawing no black balls:**
- The only way to draw no black balls is to draw 3 white balls.
- The probability of drawing 3 white balls can be calculated using combinations:
- P(No black balls) = (Number of ways to choose 3 white balls) / (Total ways to choose 3 balls)
- = C(7, 3) / C(12, 3)

3. **Calculate the combinations:**
- C(7, 3) = 7! / (3!(7-3)!) = 35
- C(12, 3) = 12! / (3!(12-3)!) = 220

4. **Calculate the probability:**
- P(No black balls) = 35 / 220 = 0.1591 (approximately)

5. **Calculate the probability of getting at least one black ball:**
- P(At least 1 black ball) = 1 - P(No black balls)
- = 1 - 0.1591 = 0.8409 (approximately)

### Quantity III:
We need to find the probability that Nirmal and Puja will contradict each other when stating the same fact.

1. **Identify the probabilities of truth-telling:**
- Nirmal tells the truth 4 out of 5 times: P(Nirmal tells the truth) = 4/5
- Puja tells the truth 6 out of 7 times: P(Puja tells the truth) = 6/7

2. **Calculate the probabilities of lying:**
- P(Nirmal lies) = 1 - P(Nirmal tells the truth) = 1/5
- P(Puja lies) = 1 - P(Puja tells the truth) = 1/7

3. **Calculate the probability of contradiction:**
- They contradict each other if:
- Nirmal tells the truth and Puja lies, or
- Nirmal lies and Puja tells the truth.
- P(Contradiction) = P(Nirmal tells the truth) * P(Puja lies) + P(Nirmal lies) * P(Puja tells the truth)
- = (4/5 * 1/7) + (1/5 * 6/7)
- = (4/35) + (6/35) = 10/35 = 2/7 ≈ 0.2857

### Summary of Results:
- Quantity I: 0.91
- Quantity II: 0.8409
- Quantity III: 0.2857

### Conclusion:
Comparing the quantities:
- Quantity I (0.91) > Quantity II (0.8409) > Quantity III (0.2857)