Question

Two fruits are randomly selected without replacement. Calculate the probability of selecting an apple and a banana.(1 point)
Responses

21190
21
190
Start Fraction 21 over 190 End Fraction

35
3
5
Start Fraction 3 over 5 End Fraction

776
7
76
Start Fraction 7 over 76 End Fraction

780 7 80 Start Fraction 7 over 80 End Fraction

Answer 1

To calculate the probability of selecting an apple and a banana when two fruits are selected randomly without replacement, we need to know the total number of fruits and how many of each fruit there are. Since you haven't provided that information, I'll assume a common example where there are apples and bananas.

Let's say there are:
- A total of 10 fruits: 4 apples and 6 bananas.

### Step 1: Calculate the total number of ways to choose 2 fruits from 10
Using the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\) where \(n\) is the total number of items, and \(k\) is the number of items to choose:

\[
C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45
\]

### Step 2: Calculate the number of ways to choose 1 apple and 1 banana
We can choose 1 apple from the 4 apples and 1 banana from the 6 bananas:

\[
C(4, 1) \times C(6, 1) = 4 \times 6 = 24
\]

### Step 3: Calculate the probability
The probability \(P\) of selecting 1 apple and 1 banana is given by the ratio of the favorable outcomes to the total outcomes:

\[
P(\text{1 apple and 1 banana}) = \frac{\text{Number of ways to choose 1 apple and 1 banana}}{\text{Total ways to choose 2 fruits}} = \frac{24}{45}
\]

### Step 4: Simplify the fraction
\[
\frac{24}{45} = \frac{8}{15}
\]

This fraction is not one of your options but demonstrates how to calculate the scenario.

Now, based on the options you have provided, none of these specific answers seem to directly correspond to the calculations we performed without specific numbers of apples and bananas.

If you have specific numbers for apples and bananas, please provide them, so I can help you find the correct probability based on that data.