A bag contains 21 markers with numbers 1 to 21. A maker is drawn at random and then replaced and then a second marker is drawn. The probability that the first number is odd and the second is even is

Options

A. `(10)/(21)`
B. `(11)/(21)`
C. `(110)/(441)`
D. `(100)/(441)`

Answer

To solve the problem, we need to find the probability that the first marker drawn is odd and the second marker drawn is even. Here’s a step-by-step solution:

### Step 1: Identify the total number of markers
The bag contains markers numbered from 1 to 21. Therefore, the total number of markers is:
\[ N = 21 \]

### Step 2: Count the odd and even markers
- Odd markers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 (Total = 11)
- Even markers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (Total = 10)

### Step 3: Calculate the probability of drawing an odd marker first
The probability \( P(\text{Odd}) \) of drawing an odd marker first is given by the ratio of the number of odd markers to the total number of markers:
\[ P(\text{Odd}) = \frac{\text{Number of odd markers}}{\text{Total markers}} = \frac{11}{21} \]

### Step 4: Calculate the probability of drawing an even marker second
Since the first marker is replaced, the total number of markers remains the same for the second draw. The probability \( P(\text{Even}) \) of drawing an even marker second is:
\[ P(\text{Even}) = \frac{\text{Number of even markers}}{\text{Total markers}} = \frac{10}{21} \]

### Step 5: Calculate the combined probability
The events of drawing the first marker and the second marker are independent since the first marker is replaced. Therefore, the combined probability \( P(\text{Odd and Even}) \) is:
\[ P(\text{Odd and Even}) = P(\text{Odd}) \times P(\text{Even}) = \left(\frac{11}{21}\right) \times \left(\frac{10}{21}\right) \]
\[ P(\text{Odd and Even}) = \frac{11 \times 10}{21 \times 21} = \frac{110}{441} \]

### Step 6: Simplify the probability
To simplify \( \frac{110}{441} \), we check for common factors. Since 110 and 441 do not have any common factors, this fraction is already in its simplest form.

### Final Answer:
The probability that the first number is odd and the second is even is:
\[ \frac{110}{441} \]

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