To find the probability of selecting one apple and one banana from the basket of fruits, we first need to determine the total number of fruits and then calculate the combinations of selecting the desired fruits.
Total Fruits
The total number of fruits is:
- Bananas: 5
- Mangoes: 8
- Apples: 7
Total number of fruits = 5 + 8 + 7 = 20
Favorable Outcomes
We can either select an apple first and then a banana, or a banana first and then an apple. We will calculate both cases and then add the probabilities together.
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Case 1: Selecting an apple first then a banana:
- Probability of selecting an apple first: \[ P(\text{Apple first}) = \frac{7}{20} \]
- After selecting an apple, there are now 19 fruits left (5 bananas, 8 mangoes, 6 apples).
- Probability of selecting a banana second: \[ P(\text{Banana second | Apple first}) = \frac{5}{19} \]
- Combined probability for this case: \[ P(\text{Apple first, then Banana}) = \frac{7}{20} \times \frac{5}{19} = \frac{35}{380} \]
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Case 2: Selecting a banana first then an apple:
- Probability of selecting a banana first: \[ P(\text{Banana first}) = \frac{5}{20} \]
- After selecting a banana, there are now 19 fruits left (4 bananas, 8 mangoes, 7 apples).
- Probability of selecting an apple second: \[ P(\text{Apple second | Banana first}) = \frac{7}{19} \]
- Combined probability for this case: \[ P(\text{Banana first, then Apple}) = \frac{5}{20} \times \frac{7}{19} = \frac{35}{380} \]
Total Probability
Adding the probabilities from both cases together: \[ P(\text{One Apple and One Banana}) = P(\text{Apple first, then Banana}) + P(\text{Banana first, then Apple}) = \frac{35}{380} + \frac{35}{380} = \frac{70}{380} \]
Now we simplify \(\frac{70}{380}\): \[ \frac{70 \div 10}{380 \div 10} = \frac{7}{38} \]
Conclusion
The probability of selecting one apple and one banana is \(\frac{7}{38}\).
Since none of the given responses are \(\frac{7}{38}\), please verify, but as per our calculations, that is the correct answer.