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

776
Start Fraction 7 over 76 End Fraction

780
Start Fraction 7 over 80 End Fraction

21190
Start Fraction 21 over 190 End Fraction

35

Answer 1

To calculate the probability of selecting one apple and one banana from the basket, we first look at the total number of fruits and the combinations possible.

### Step 1: Total fruits
The basket contains:
- 5 bananas
- 8 mangoes
- 7 apples

Total number of fruits = 5 + 8 + 7 = 20 fruits.

### Step 2: Calculate the total ways to select 2 fruits
The number of ways to select 2 fruits from 20 can be calculated using the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n - r)!}
\]
So, the total ways to choose 2 fruits from 20 is:
\[
\binom{20}{2} = \frac{20!}{2!(20 - 2)!} = \frac{20 \times 19}{2 \times 1} = 190.
\]

### Step 3: Calculate the ways to select 1 apple and 1 banana
The number of ways to select 1 apple from 7 is:
\[
\binom{7}{1} = 7.
\]
The number of ways to select 1 banana from 5 is:
\[
\binom{5}{1} = 5.
\]
Thus, the total ways to select 1 apple and 1 banana is:
\[
7 \times 5 = 35.
\]

### Step 4: Calculate the probability
Now, we can find the probability of selecting 1 apple and 1 banana:
\[
P(\text{1 apple and 1 banana}) = \frac{\text{Number of ways to select 1 apple and 1 banana}}{\text{Total ways to select 2 fruits}} = \frac{35}{190}.
\]

### Step 5: Simplify the fraction
To simplify \( \frac{35}{190} \):
\[
\frac{35}{190} = \frac{7}{38}.
\]

### Conclusion
The final probability of selecting one apple and one banana is:
\[
\frac{7}{38}.
\]

However, from the given choices, none directly reflect this. Let's check the possible choices as and determine if they might need approximation or have simplified forms known.

Among your provided options, the closest and most logical answer based on selection indicates \( \frac{21}{190} \) might be mistakenly assessed in terms of pairs' approach or mixed categorically. Ensure each proportion adequately assigns weight to the various fruits to backtrack to single selections if the aggregate led us thus far.

It depends on the instructional accuracy intended. The true mathematical approach through strict construction indicates this—but verify the lesson construction on mixed selections particularly weighted without replacement to clarify further drops per alternate methods prescribed above if (inputs) warrant a look at straightforward. But again, mathematical representation holds firm to \( \frac{7}{38} \).

Confirm final notation from involved instructional fragments or broader method prompts in classroom communication!