Question
Ryan has a pack of 10 pens. 3 pens are red, 4 are black, and 3 are blue. What are the chances that he pulls out a red pen and a black pen? Assume that the pens are pulled without replacement. Round to the nearest whole number.
Answers
GPT-4o mini
To find the probability that Ryan pulls out a red pen and a black pen from his pack of 10 pens (3 red, 4 black, and 3 blue), we can break the problem down into two cases: pulling out a red pen first and then a black pen, or pulling out a black pen first and then a red pen.
1. **Case 1: Pulling a red pen first, then a black pen**
- The probability of pulling out a red pen first:
\[
P(\text{red first}) = \frac{3 \text{ red pens}}{10 \text{ total pens}} = \frac{3}{10}
\]
- After pulling a red pen, there are 9 pens left (2 red, 4 black, 3 blue). The probability of then pulling out a black pen:
\[
P(\text{black second | red first}) = \frac{4 \text{ black pens}}{9 \text{ remaining pens}} = \frac{4}{9}
\]
- Therefore, the combined probability for this case is:
\[
P(\text{red first, black second}) = P(\text{red first}) \times P(\text{black second | red first}) = \frac{3}{10} \times \frac{4}{9} = \frac{12}{90} = \frac{2}{15}
\]
2. **Case 2: Pulling a black pen first, then a red pen**
- The probability of pulling out a black pen first:
\[
P(\text{black first}) = \frac{4 \text{ black pens}}{10 \text{ total pens}} = \frac{4}{10} = \frac{2}{5}
\]
- After pulling a black pen, there are 9 pens left (3 red, 3 black, 3 blue). The probability of then pulling out a red pen:
\[
P(\text{red second | black first}) = \frac{3 \text{ red pens}}{9 \text{ remaining pens}} = \frac{3}{9} = \frac{1}{3}
\]
- Therefore, the combined probability for this case is:
\[
P(\text{black first, red second}) = P(\text{black first}) \times P(\text{red second | black first}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}
\]
3. **Overall Probability**
- Now, we add the probabilities from both cases to find the total probability of pulling out one red pen and one black pen regardless of the order:
\[
P(\text{red and black in any order}) = P(\text{red first, black second}) + P(\text{black first, red second}) = \frac{2}{15} + \frac{2}{15} = \frac{4}{15}
\]
4. **Final Calculation: Rounding**
- The probability \( \frac{4}{15} \) can be approximated as a decimal for rounding:
\[
\frac{4}{15} \approx 0.2667
\]
- To convert this to a percentage:
\[
0.2667 \times 100 \approx 26.67\%
\]
- Rounding to the nearest whole number, we get:
\[
\boxed{27}
\]
Therefore, the chances that Ryan pulls out a red pen and a black pen (in any order) is approximately **27%**.
1. **Case 1: Pulling a red pen first, then a black pen**
- The probability of pulling out a red pen first:
\[
P(\text{red first}) = \frac{3 \text{ red pens}}{10 \text{ total pens}} = \frac{3}{10}
\]
- After pulling a red pen, there are 9 pens left (2 red, 4 black, 3 blue). The probability of then pulling out a black pen:
\[
P(\text{black second | red first}) = \frac{4 \text{ black pens}}{9 \text{ remaining pens}} = \frac{4}{9}
\]
- Therefore, the combined probability for this case is:
\[
P(\text{red first, black second}) = P(\text{red first}) \times P(\text{black second | red first}) = \frac{3}{10} \times \frac{4}{9} = \frac{12}{90} = \frac{2}{15}
\]
2. **Case 2: Pulling a black pen first, then a red pen**
- The probability of pulling out a black pen first:
\[
P(\text{black first}) = \frac{4 \text{ black pens}}{10 \text{ total pens}} = \frac{4}{10} = \frac{2}{5}
\]
- After pulling a black pen, there are 9 pens left (3 red, 3 black, 3 blue). The probability of then pulling out a red pen:
\[
P(\text{red second | black first}) = \frac{3 \text{ red pens}}{9 \text{ remaining pens}} = \frac{3}{9} = \frac{1}{3}
\]
- Therefore, the combined probability for this case is:
\[
P(\text{black first, red second}) = P(\text{black first}) \times P(\text{red second | black first}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}
\]
3. **Overall Probability**
- Now, we add the probabilities from both cases to find the total probability of pulling out one red pen and one black pen regardless of the order:
\[
P(\text{red and black in any order}) = P(\text{red first, black second}) + P(\text{black first, red second}) = \frac{2}{15} + \frac{2}{15} = \frac{4}{15}
\]
4. **Final Calculation: Rounding**
- The probability \( \frac{4}{15} \) can be approximated as a decimal for rounding:
\[
\frac{4}{15} \approx 0.2667
\]
- To convert this to a percentage:
\[
0.2667 \times 100 \approx 26.67\%
\]
- Rounding to the nearest whole number, we get:
\[
\boxed{27}
\]
Therefore, the chances that Ryan pulls out a red pen and a black pen (in any order) is approximately **27%**.