Question

A basket of fruits 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.(1 point)
Responses

780
Start Fraction 7 over 80 End Fraction

35
Start Fraction 3 over 5 End Fraction

776
Start Fraction 7 over 76 End Fraction

21/190

Answer 1

To calculate the probability of selecting one apple and one banana from the basket of fruits, we will follow these steps:

1. **Total number of fruits:**
There are 5 bananas, 8 mangoes, and 7 apples.
Total fruits = 5 + 8 + 7 = 20.

2. **Ways to select 1 apple and 1 banana:**
- The number of ways to choose an apple = 7 (since there are 7 apples).
- The number of ways to choose a banana = 5 (since there are 5 bananas).
- The number of ways to choose 1 apple and 1 banana in any order = \(7 \times 5 = 35\).

3. **Total ways to select 2 fruits from 20:**
We can calculate this using the combination formula \(C(n, r)\), where \(n\) is the total number of items (fruits) and \(r\) is the number of items to choose.
Total ways to choose 2 fruits from 20 = \(C(20, 2) = \frac{20 \times 19}{2} = 190\).

4. **Probability of selecting 1 apple and 1 banana:**
Probability = (Number of favorable outcomes) / (Total outcomes)
Probability = \( \frac{35}{190} \).

Now, we can simplify \( \frac{35}{190} \):
- Both 35 and 190 can be divided by 5.
- \( \frac{35 \div 5}{190 \div 5} = \frac{7}{38} \).

Thus, the probability of selecting one apple and one banana is \( \frac{7}{38} \).

None of the responses provided in the original question directly matches \( \frac{7}{38} \), which might indicate an error in the provided responses. Please verify the question or the options as necessary.