10 packets of suger each marked 1 kg actually contained the following weights 0.6 kg, 1.050 kg, 1.1 kg, -0.98 kg, 0.92 kg, 1.3 kg, 1.4 kg, 1.00 kg, 0.94 kg, 1.03 kg. Out of these packets, one packet is chosen at random, what is the probability that the chosen packet contain
(i) more than 1 kg sugar,
(ii) 1 kg sugar,
(iii) less than 1 kg sugar?

Answer

To solve the problem, we need to determine the probability of selecting a packet of sugar that contains specific amounts of sugar. We will categorize the packets based on their weights and then calculate the probabilities for each category.

### Given Weights of Sugar Packets:
1. 0.6 kg
2. 1.050 kg
3. 1.1 kg
4. 0.98 kg
5. 0.92 kg
6. 1.3 kg
7. 1.4 kg
8. 1.00 kg
9. 0.94 kg
10. 1.03 kg

### Step 1: Count the Total Number of Packets
The total number of packets is 10.

### Step 2: Categorize the Packets
We will categorize the packets into three groups based on their weights:

1. **More than 1 kg**:
- Packets: 1.1 kg, 1.3 kg, 1.4 kg, 1.050 kg, 1.03 kg
- Count: 5 packets

2. **Exactly 1 kg**:
- Packet: 1.00 kg
- Count: 1 packet

3. **Less than 1 kg**:
- Packets: 0.6 kg, 0.98 kg, 0.92 kg, 0.94 kg
- Count: 4 packets

### Step 3: Calculate the Probabilities
Now we will calculate the probabilities for each category.

#### (i) Probability of selecting a packet that contains **more than 1 kg** of sugar:
- Favorable outcomes = 5 (packets with weights more than 1 kg)
- Total outcomes = 10

\[ P(\text{more than 1 kg}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{10} = \frac{1}{2} \]

#### (ii) Probability of selecting a packet that contains **exactly 1 kg** of sugar:
- Favorable outcomes = 1 (packet with weight exactly 1 kg)
- Total outcomes = 10

\[ P(\text{exactly 1 kg}) = \frac{1}{10} \]

#### (iii) Probability of selecting a packet that contains **less than 1 kg** of sugar:
- Favorable outcomes = 4 (packets with weights less than 1 kg)
- Total outcomes = 10

\[ P(\text{less than 1 kg}) = \frac{4}{10} = \frac{2}{5} \]

### Final Answers:
- (i) Probability of more than 1 kg: \(\frac{1}{2}\)
- (ii) Probability of exactly 1 kg: \(\frac{1}{10}\)
- (iii) Probability of less than 1 kg: \(\frac{2}{5}\)