A bag contains 17 markers with numbers 1 to 17 . A marker is drawn at random and then replaced, a second marker is drawn then the probability that first number is even and second is odd, is

Options

A. `(72)/(712)`
B. `(72)/(289)`
C. `(.^(8)C_(1)xx.^(0)C_(1))/(.^(17)C_(2))`
D. `(.^(8)C_(1)xx.^(9)C_(1))/((.^(17)C_(1))^2`.

Answer

To solve the problem, we need to find the probability that the first marker drawn is even and the second marker drawn is odd. Let's break this down step by step.

### Step 1: Identify the total number of markers
The bag contains markers numbered from 1 to 17. Therefore, the total number of markers is:
\[ \text{Total markers} = 17 \]

### Step 2: Count the even and odd markers
The even markers in the range from 1 to 17 are:
\[ 2, 4, 6, 8, 10, 12, 14, 16 \]
This gives us a total of 8 even markers.

The odd markers in the range from 1 to 17 are:
\[ 1, 3, 5, 7, 9, 11, 13, 15, 17 \]
This gives us a total of 9 odd markers.

### Step 3: Calculate the probability of drawing an even marker first
The probability of drawing an even marker on the first draw is given by the ratio of the number of even markers to the total number of markers:
\[ P(\text{Even first}) = \frac{\text{Number of even markers}}{\text{Total markers}} = \frac{8}{17} \]

### Step 4: Calculate the probability of drawing an odd marker second
Since the first marker is replaced, the total number of markers remains the same for the second draw. The probability of drawing an odd marker on the second draw is given by:
\[ P(\text{Odd second}) = \frac{\text{Number of odd markers}}{\text{Total markers}} = \frac{9}{17} \]

### Step 5: Calculate the combined probability
Since the two events (drawing an even marker first and an odd marker second) are independent (due to replacement), we can multiply their probabilities:
\[ P(\text{Even first and Odd second}) = P(\text{Even first}) \times P(\text{Odd second}) \]
Substituting the probabilities we calculated:
\[ P(\text{Even first and Odd second}) = \frac{8}{17} \times \frac{9}{17} = \frac{72}{289} \]

### Final Answer
Thus, the probability that the first number is even and the second number is odd is:
\[ \frac{72}{289} \]

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