Question

a basket of fruits contains five bananas, eight mangoes and seven apples. Two fruits are randomly selected without replacement. Calculate probability of selecting an apple and a banana.

Answer 1

To calculate the probability of selecting an apple and a banana from the basket of fruits, we can follow these steps:

1. **Total number of fruits**:
- Bananas: 5
- Mangoes: 8
- Apples: 7
- Total = 5 + 8 + 7 = 20 fruits

2. **Total combinations of selecting 2 fruits**:
The total number of ways to choose 2 fruits from 20 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of fruits and \( r \) is the number of fruits to choose.

\[
\text{Total combinations} = \binom{20}{2} = \frac{20 \times 19}{2 \times 1} = 190
\]

3. **Calculate the combinations for selecting 1 apple and 1 banana**:
- Ways to choose 1 apple from 7: \( \binom{7}{1} = 7 \)
- Ways to choose 1 banana from 5: \( \binom{5}{1} = 5 \)

Therefore, the total combinations for selecting 1 apple and 1 banana is:

\[
\text{Combinations for 1 apple and 1 banana} = 7 \times 5 = 35
\]

4. **Calculate the probability**:
To find the probability of selecting 1 apple and 1 banana, we divide the combinations of selecting 1 apple and 1 banana by the total combinations of selecting 2 fruits:

\[
\text{Probability} = \frac{\text{Combinations for 1 apple and 1 banana}}{\text{Total combinations}} = \frac{35}{190}
\]

5. **Simplifying the probability**:
To simplify \( \frac{35}{190} \), we can divide both the numerator and the denominator by 5:

\[
\frac{35 \div 5}{190 \div 5} = \frac{7}{38}
\]

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