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A bag contains 17 markers with numbers 1...

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

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`.

Text Solution

AI Generated Solution

The correct Answer is:
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 it down step by step. ### Step 1: Identify the total numbers and categorize them The bag contains markers numbered from 1 to 17. We need to determine how many of these markers are even and how many are odd. - **Even Numbers**: The even numbers from 1 to 17 are 2, 4, 6, 8, 10, 12, 14, 16. This gives us a total of **8 even numbers**. - **Odd Numbers**: The odd numbers from 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17. This gives us a total of **9 odd numbers**. ### Step 2: Calculate the probability of drawing an even number first The probability of drawing an even number first can be calculated using the formula: \[ P(\text{Even}) = \frac{\text{Number of Even Numbers}}{\text{Total Number of Markers}} = \frac{8}{17} \] ### Step 3: Calculate the probability of drawing an odd number 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 number second is: \[ P(\text{Odd}) = \frac{\text{Number of Odd Numbers}}{\text{Total Number of Markers}} = \frac{9}{17} \] ### Step 4: Calculate the combined probability To find the probability that the first number is even and the second number is odd, we multiply the probabilities from Step 2 and Step 3: \[ P(\text{Even first and Odd second}) = P(\text{Even}) \times P(\text{Odd}) = \frac{8}{17} \times \frac{9}{17} \] ### Step 5: Simplify the expression Now we simplify the expression: \[ P(\text{Even first and Odd second}) = \frac{8 \times 9}{17 \times 17} = \frac{72}{289} \] ### Final Answer Thus, the probability that the first marker drawn is even and the second marker drawn is odd is: \[ \frac{72}{289} \] ---
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