A jar contains 3 red and 2 white marbles. Two marbles are picked without replacement.
`{:("Quantity A","Quantity B"),("the probability of picking two","The probability of picking exactly"),("red marbles","one red and one white marble"):}`

Answer

To solve the problem, we need to calculate the probabilities for both Quantity A and Quantity B.

### Step 1: Calculate the Probability of Picking Two Red Marbles (Quantity A)

1. **Total Marbles**: There are 3 red marbles and 2 white marbles, making a total of 5 marbles.
2. **Probability of Picking the First Red Marble**: The probability of picking the first red marble is given by the ratio of red marbles to total marbles:
\[ P(\text{First Red}) = \frac{3}{5} \]
3. **Probability of Picking the Second Red Marble**: After picking the first red marble, there are now 2 red marbles left and a total of 4 marbles remaining:
\[ P(\text{Second Red | First Red}) = \frac{2}{4} = \frac{1}{2} \]
4. **Combined Probability**: The combined probability of both events (picking two red marbles) is the product of the two probabilities:
\[ P(\text{Two Red}) = P(\text{First Red}) \times P(\text{Second Red | First Red}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10} = 0.3 \]

### Step 2: Calculate the Probability of Picking One Red and One White Marble (Quantity B)

1. **Case 1: First Red, Second White**:
- Probability of picking the first red marble:
\[ P(\text{First Red}) = \frac{3}{5} \]
- Probability of picking the second white marble (after picking one red):
\[ P(\text{Second White | First Red}) = \frac{2}{4} = \frac{1}{2} \]
- Combined probability for this case:
\[ P(\text{Red then White}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10} \]

2. **Case 2: First White, Second Red**:
- Probability of picking the first white marble:
\[ P(\text{First White}) = \frac{2}{5} \]
- Probability of picking the second red marble (after picking one white):
\[ P(\text{Second Red | First White}) = \frac{3}{4} \]
- Combined probability for this case:
\[ P(\text{White then Red}) = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10} \]

3. **Total Probability for Quantity B**: Add the probabilities of both cases:
\[ P(\text{One Red and One White}) = P(\text{Red then White}) + P(\text{White then Red}) = \frac{3}{10} + \frac{3}{10} = \frac{6}{10} = 0.6 \]

### Step 3: Compare the Quantities

- **Quantity A**: Probability of picking two red marbles = 0.3
- **Quantity B**: Probability of picking one red and one white marble = 0.6

Since \(0.6 > 0.3\), we conclude that Quantity B is greater than Quantity A.

### Final Answer
**Quantity B is higher than Quantity A.**

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