Question
A basket of fruit contains 5 bananas, 8 mangoes, and 7 apples. Two fruits are randomly selected without replacement. Calculate the probability of selecting an apple and a banana.
a. 7/76
b. 3/5
c. 7/80
d. 21/190
a. 7/76
b. 3/5
c. 7/80
d. 21/190
Answers
GPT-4o mini
To calculate the probability of selecting an apple and a banana from the basket, we will first determine the total number of fruits and then find the number of favorable outcomes.
1. **Total fruits in the basket**:
\[
\text{Total fruits} = \text{Bananas} + \text{Mangoes} + \text{Apples} = 5 + 8 + 7 = 20
\]
2. **Select an apple and then a banana**:
- Probability of selecting an apple first:
\[
P(A) = \frac{\text{Number of apples}}{\text{Total fruits}} = \frac{7}{20}
\]
- After selecting an apple, there are now 19 fruits left and the number of bananas remains 5. Therefore, the probability of selecting a banana next:
\[
P(B|A) = \frac{\text{Number of bananas}}{\text{Total fruits left}} = \frac{5}{19}
\]
- Thus, the combined probability of this scenario (selecting an apple first and then a banana) is:
\[
P(A \text{ and then } B) = P(A) \cdot P(B|A) = \frac{7}{20} \cdot \frac{5}{19} = \frac{35}{380}
\]
3. **Select a banana and then an apple**:
- Probability of selecting a banana first:
\[
P(B) = \frac{5}{20} = \frac{1}{4}
\]
- After selecting a banana, there are now 19 fruits left and the number of apples remains 7. Therefore, the probability of selecting an apple next:
\[
P(A|B) = \frac{7}{19}
\]
- Thus, the combined probability of this scenario (selecting a banana first and then an apple) is:
\[
P(B \text{ and then } A) = P(B) \cdot P(A|B) = \frac{1}{4} \cdot \frac{7}{19} = \frac{7}{76}
\]
4. **Combined probability of selecting an apple and a banana in any order**:
\[
P(\text{Apple and Banana}) = P(A \text{ then } B) + P(B \text{ then } A) = \frac{35}{380} + \frac{7}{76}
\]
- Convert \(\frac{7}{76}\) to a common denominator of 380:
\[
\frac{7}{76} = \frac{7 \times 5}{76 \times 5} = \frac{35}{380}
\]
- Thus, add them:
\[
P(\text{Apple and Banana}) = \frac{35}{380} + \frac{35}{380} = \frac{70}{380}
\]
- Simplifying \(\frac{70}{380}\):
\[
\frac{70 \div 10}{380 \div 10} = \frac{7}{38}
\]
Finally, the probability of selecting an apple and a banana is:
\[
\text{Probability} = \frac{7}{38}
\]
After reviewing the choices again and checking all formulations, it appears that none of the options listed in the question match \(\frac{7}{38}\). However, based on the work done, the calculations have been verified.
If we check the four options again:
- None of the given options are correct.
**Conclusion**: The correct probability of selecting one apple and one banana is \(\frac{7}{38}\), which does not match any of the choices provided.
1. **Total fruits in the basket**:
\[
\text{Total fruits} = \text{Bananas} + \text{Mangoes} + \text{Apples} = 5 + 8 + 7 = 20
\]
2. **Select an apple and then a banana**:
- Probability of selecting an apple first:
\[
P(A) = \frac{\text{Number of apples}}{\text{Total fruits}} = \frac{7}{20}
\]
- After selecting an apple, there are now 19 fruits left and the number of bananas remains 5. Therefore, the probability of selecting a banana next:
\[
P(B|A) = \frac{\text{Number of bananas}}{\text{Total fruits left}} = \frac{5}{19}
\]
- Thus, the combined probability of this scenario (selecting an apple first and then a banana) is:
\[
P(A \text{ and then } B) = P(A) \cdot P(B|A) = \frac{7}{20} \cdot \frac{5}{19} = \frac{35}{380}
\]
3. **Select a banana and then an apple**:
- Probability of selecting a banana first:
\[
P(B) = \frac{5}{20} = \frac{1}{4}
\]
- After selecting a banana, there are now 19 fruits left and the number of apples remains 7. Therefore, the probability of selecting an apple next:
\[
P(A|B) = \frac{7}{19}
\]
- Thus, the combined probability of this scenario (selecting a banana first and then an apple) is:
\[
P(B \text{ and then } A) = P(B) \cdot P(A|B) = \frac{1}{4} \cdot \frac{7}{19} = \frac{7}{76}
\]
4. **Combined probability of selecting an apple and a banana in any order**:
\[
P(\text{Apple and Banana}) = P(A \text{ then } B) + P(B \text{ then } A) = \frac{35}{380} + \frac{7}{76}
\]
- Convert \(\frac{7}{76}\) to a common denominator of 380:
\[
\frac{7}{76} = \frac{7 \times 5}{76 \times 5} = \frac{35}{380}
\]
- Thus, add them:
\[
P(\text{Apple and Banana}) = \frac{35}{380} + \frac{35}{380} = \frac{70}{380}
\]
- Simplifying \(\frac{70}{380}\):
\[
\frac{70 \div 10}{380 \div 10} = \frac{7}{38}
\]
Finally, the probability of selecting an apple and a banana is:
\[
\text{Probability} = \frac{7}{38}
\]
After reviewing the choices again and checking all formulations, it appears that none of the options listed in the question match \(\frac{7}{38}\). However, based on the work done, the calculations have been verified.
If we check the four options again:
- None of the given options are correct.
**Conclusion**: The correct probability of selecting one apple and one banana is \(\frac{7}{38}\), which does not match any of the choices provided.