In the given question two quantities are given one as quantity -I and the other is quantity II you have to determine relationship between these two quantities and choose the appropriate options as given
Quantity-I:four persons are chosen at random from a group of 3 men ,5 women and 4 children what is probability that exactly two of them are men?
Quantity II:A box contains 3 ballons of one shape 4 ballons of another shape and 5 ballons of another shape three ballons of them are drawn at random what is probability that all three are of different shapes?

Options

A. `quantity-IgtII`
B. `quantity IltII`
C. `quantity IleII`
D. `quantity IgeII`

Answer

To solve the problem, we need to calculate the probabilities for both Quantity-I and Quantity-II step by step.

### Quantity-I: Probability of Selecting Exactly Two Men

1. **Identify Total Persons**:
- There are 3 men, 5 women, and 4 children.
- Total = 3 + 5 + 4 = 12 persons.

2. **Total Ways to Choose 4 Persons from 12**:
- The number of ways to choose 4 persons from 12 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Thus, \( C(12, 4) = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} \).
- Calculating this gives:
\[ C(12, 4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495. \]

3. **Favorable Outcomes for Exactly 2 Men**:
- We need to choose 2 men from 3 and 2 persons from the remaining 9 (5 women + 4 children).
- Ways to choose 2 men from 3: \( C(3, 2) = 3 \).
- Ways to choose 2 persons from 9: \( C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \).
- Total favorable outcomes = \( C(3, 2) \times C(9, 2) = 3 \times 36 = 108 \).

4. **Calculate Probability**:
- Probability = \( \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{108}{495} \).
- Simplifying this gives:
\[ \frac{108}{495} = \frac{12}{55} \approx 0.2182. \]

### Quantity-II: Probability of Drawing 3 Balloons of Different Shapes

1. **Identify Total Balloons**:
- There are 3 balloons of shape A, 4 of shape B, and 5 of shape C.
- Total = 3 + 4 + 5 = 12 balloons.

2. **Total Ways to Choose 3 Balloons from 12**:
- Total ways = \( C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \).

3. **Favorable Outcomes for Different Shapes**:
- We need to select 1 balloon from each shape.
- Ways to choose 1 from shape A = \( C(3, 1) = 3 \).
- Ways to choose 1 from shape B = \( C(4, 1) = 4 \).
- Ways to choose 1 from shape C = \( C(5, 1) = 5 \).
- Total favorable outcomes = \( 3 \times 4 \times 5 = 60 \).

4. **Calculate Probability**:
- Probability = \( \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{60}{220} \).
- Simplifying this gives:
\[ \frac{60}{220} = \frac{3}{11} \approx 0.2727. \]

### Conclusion

Now we compare the two probabilities:
- Quantity-I: \( \approx 0.2182 \)
- Quantity-II: \( \approx 0.2727 \)

Thus, Quantity-II is greater than Quantity-I.

### Final Answer
**Quantity II is greater than Quantity I.**