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)

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

1. Identify the total number of fruits:

   • Total fruits = 5 bananas + 8 mangoes + 7 apples = 20 fruits

2. Calculate the number of favorable outcomes for selecting 1 apple and 1 banana:

   • We can have two scenarios for selecting an apple and a banana: selecting a banana first and then an apple, or selecting an apple first and then a banana.

   Scenario 1: Select a banana first, then an apple

   • Probability of selecting a banana first: $\frac{5}{20}$
   • After selecting a banana, the number of remaining fruits: 19 (4 bananas, 8 mangoes, 7 apples)
   • Probability of selecting an apple next: $\frac{7}{19}$
   • Combined probability for this scenario: $$P(\text{banana first, then apple})=\frac{5}{20}\times \frac{7}{19}=\frac{5\times 7}{20\times 19}=\frac{35}{380}$$

   Scenario 2: Select an apple first, then a banana

   • Probability of selecting an apple first: $\frac{7}{20}$
   • After selecting an apple, the number of remaining fruits: 19 (5 bananas, 8 mangoes, 6 apples)
   • Probability of selecting a banana next: $\frac{5}{19}$
   • Combined probability for this scenario: $$P(\text{apple first, then banana})=\frac{7}{20}\times \frac{5}{19}=\frac{7\times 5}{20\times 19}=\frac{35}{380}$$

3. Total probability of selecting one apple and one banana:

   • Add the probabilities from both scenarios: $$P(\text{one apple and one banana})=P(\text{banana first, then apple})+P(\text{apple first, then banana})=\frac{35}{380}+\frac{35}{380}=\frac{70}{380}$$

4. Simplify the probability:

   • $\frac{70}{380}=\frac{7}{38}$

Therefore, the probability of selecting one apple and one banana is $\frac{7}{38}$.