Answer :
To find the probability that one marble is yellow and the other marble is red when two marbles are chosen from a bag that contains eight yellow marbles, nine green marbles, three purple marbles, and five red marbles, we need to proceed as follows:
1. Calculate the total number of marbles in the bag:
[tex]\[ \text{Total marbles} = 8 (\text{yellow}) + 9 (\text{green}) + 3 (\text{purple}) + 5 (\text{red}) = 25 \][/tex]
2. Calculate the probability of drawing one yellow marble and then one red marble:
- First, the probability of drawing a yellow marble:
[tex]\[ P(\text{Yellow first}) = \frac{8}{25} \][/tex]
- After one yellow marble is drawn, there are now 24 marbles left, including 5 red marbles. The probability of drawing a red marble next:
[tex]\[ P(\text{Red second | Yellow first}) = \frac{5}{24} \][/tex]
- Therefore, the combined probability for this scenario is:
[tex]\[ P(\text{Yellow first and Red second}) = \frac{8}{25} \times \frac{5}{24} \][/tex]
3. Calculate the probability of drawing one red marble and then one yellow marble:
- First, the probability of drawing a red marble:
[tex]\[ P(\text{Red first}) = \frac{5}{25} \][/tex]
- After one red marble is drawn, there are now 24 marbles left, including 8 yellow marbles. The probability of drawing a yellow marble next:
[tex]\[ P(\text{Yellow second | Red first}) = \frac{8}{24} \][/tex]
- Therefore, the combined probability for this scenario is:
[tex]\[ P(\text{Red first and Yellow second}) = \frac{5}{25} \times \frac{8}{24} \][/tex]
4. Combine both scenarios:
Since the two events are mutually exclusive (they don't overlap), the overall probability is the sum of both combined probabilities:
[tex]\[ P(\text{One Yellow and One Red}) = P(\text{Yellow first and Red second}) + P(\text{Red first and Yellow second}) \][/tex]
[tex]\[ P(\text{One Yellow and One Red}) = \left( \frac{8}{25} \times \frac{5}{24} \right) + \left( \frac{5}{25} \times \frac{8}{24} \right) \][/tex]
5. Simplify the expression:
Each term in the sum simplifies to:
[tex]\[ \frac{8 \times 5}{25 \times 24} + \frac{5 \times 8}{25 \times 24} = 2 \times \frac{8 \times 5}{25 \times 24} = 2 \times \frac{40}{600} = 2 \times \frac{1}{15} = \frac{2}{15} \][/tex]
Therefore, the expression that gives the probability that one marble is yellow and the other marble is red is:
[tex]\[ P(\text{Yellow and Red}) = \frac{2}{15} \][/tex]
Given the options provided in the original question, the closest representation of this result is:
[tex]\[ P(\text{Y and R}) = \frac{\left(C_1 C_8 \right) \left(C_1 \right)}{{ }_2 C_{25}} \][/tex]
This matches with the simplification process described and indicates the correct answer is:
[tex]\[ \mathbb{P}(Y \text{ and } R) = \frac{8}{25} \times \frac{5}{24} + \frac{5}{25} \times \frac{8}{24} = \left(\frac{8}{25} \cdot \frac{5}{24}\right) + \left(\frac{5}{25} \cdot \frac{8}{24}\right) = 2 \left(\frac{8}{25} \cdot \frac{5}{24}\right) = \frac{2}{15} \][/tex]
1. Calculate the total number of marbles in the bag:
[tex]\[ \text{Total marbles} = 8 (\text{yellow}) + 9 (\text{green}) + 3 (\text{purple}) + 5 (\text{red}) = 25 \][/tex]
2. Calculate the probability of drawing one yellow marble and then one red marble:
- First, the probability of drawing a yellow marble:
[tex]\[ P(\text{Yellow first}) = \frac{8}{25} \][/tex]
- After one yellow marble is drawn, there are now 24 marbles left, including 5 red marbles. The probability of drawing a red marble next:
[tex]\[ P(\text{Red second | Yellow first}) = \frac{5}{24} \][/tex]
- Therefore, the combined probability for this scenario is:
[tex]\[ P(\text{Yellow first and Red second}) = \frac{8}{25} \times \frac{5}{24} \][/tex]
3. Calculate the probability of drawing one red marble and then one yellow marble:
- First, the probability of drawing a red marble:
[tex]\[ P(\text{Red first}) = \frac{5}{25} \][/tex]
- After one red marble is drawn, there are now 24 marbles left, including 8 yellow marbles. The probability of drawing a yellow marble next:
[tex]\[ P(\text{Yellow second | Red first}) = \frac{8}{24} \][/tex]
- Therefore, the combined probability for this scenario is:
[tex]\[ P(\text{Red first and Yellow second}) = \frac{5}{25} \times \frac{8}{24} \][/tex]
4. Combine both scenarios:
Since the two events are mutually exclusive (they don't overlap), the overall probability is the sum of both combined probabilities:
[tex]\[ P(\text{One Yellow and One Red}) = P(\text{Yellow first and Red second}) + P(\text{Red first and Yellow second}) \][/tex]
[tex]\[ P(\text{One Yellow and One Red}) = \left( \frac{8}{25} \times \frac{5}{24} \right) + \left( \frac{5}{25} \times \frac{8}{24} \right) \][/tex]
5. Simplify the expression:
Each term in the sum simplifies to:
[tex]\[ \frac{8 \times 5}{25 \times 24} + \frac{5 \times 8}{25 \times 24} = 2 \times \frac{8 \times 5}{25 \times 24} = 2 \times \frac{40}{600} = 2 \times \frac{1}{15} = \frac{2}{15} \][/tex]
Therefore, the expression that gives the probability that one marble is yellow and the other marble is red is:
[tex]\[ P(\text{Yellow and Red}) = \frac{2}{15} \][/tex]
Given the options provided in the original question, the closest representation of this result is:
[tex]\[ P(\text{Y and R}) = \frac{\left(C_1 C_8 \right) \left(C_1 \right)}{{ }_2 C_{25}} \][/tex]
This matches with the simplification process described and indicates the correct answer is:
[tex]\[ \mathbb{P}(Y \text{ and } R) = \frac{8}{25} \times \frac{5}{24} + \frac{5}{25} \times \frac{8}{24} = \left(\frac{8}{25} \cdot \frac{5}{24}\right) + \left(\frac{5}{25} \cdot \frac{8}{24}\right) = 2 \left(\frac{8}{25} \cdot \frac{5}{24}\right) = \frac{2}{15} \][/tex]
