To calculate the chances of Ryan pulling out a red pen and a black pen from a pack of 10 pens without replacement, we can break the problem down into two scenarios: first, he pulls out a red pen and then he pulls out a black pen, and second, he pulls out a black pen followed by a red pen.
Scenario 1: Red pen first, then black pen
- The probability of picking a red pen first: \[ P(\text{Red first}) = \frac{3}{10} \]
- After pulling out a red pen, there are 9 pens left, including 4 black pens. The probability of now picking a black pen: \[ P(\text{Black second | Red first}) = \frac{4}{9} \]
- Therefore, the combined probability for this scenario: \[ P(\text{Red first, Black second}) = P(\text{Red first}) \times P(\text{Black second | Red first}) = \frac{3}{10} \times \frac{4}{9} = \frac{12}{90} \]
Scenario 2: Black pen first, then red pen
- The probability of picking a black pen first: \[ P(\text{Black first}) = \frac{4}{10} \]
- After pulling out a black pen, there are still 3 red pens among the remaining 9 pens. The probability of now picking a red pen: \[ P(\text{Red second | Black first}) = \frac{3}{9} \]
- Therefore, the combined probability for this scenario: \[ P(\text{Black first, Red second}) = P(\text{Black first}) \times P(\text{Red second | Black first}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} \]
Total Probability
Now, we can add the two probabilities together: \[ P(\text{Red and Black}) = P(\text{Red first, Black second}) + P(\text{Black first, Red second}) = \frac{12}{90} + \frac{12}{90} = \frac{24}{90} \] This simplifies to: \[ P(\text{Red and Black}) = \frac{4}{15} \]
Convert to Percentage
To find the percentage, we multiply by 100: \[ \frac{4}{15} \approx 0.2667 \] So as a percentage: \[ 0.2667 \times 100 \approx 26.67% \] Rounding to the nearest whole number, the chances are: \[ \boxed{27%} \]