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

7/76
Start Fraction 7 over 76 End Fraction

3/5
  Start Fraction 3 over 5 End Fraction

21/190
  Start Fraction 21 over 190 End Fraction

7/80

Choose the answer

1 answer

To calculate the probability of selecting one apple and one banana from the basket, we’ll first need to find the total number of fruits and then consider two scenarios: selecting an apple first or selecting a banana first.

Total fruits in the basket:

  • Bananas: 5
  • Mangoes: 8
  • Apples: 7

Total fruits = 5 + 8 + 7 = 20 fruits

Scenario 1: Selecting an apple first, then a banana

  1. Probability of selecting an apple first: \[ P(\text{Apple first}) = \frac{7}{20} \]
  2. After selecting an apple, we have:
    • 5 bananas
    • 8 mangoes
    • 6 apples (since we have taken 1 apple)

Total fruits left = 20 - 1 = 19

  1. Probability of selecting a banana next: \[ P(\text{Banana second | Apple first}) = \frac{5}{19} \]

So, the probability of selecting an apple first and then a banana is: \[ P(\text{Apple, then Banana}) = P(\text{Apple first}) \times P(\text{Banana second | Apple first}) = \frac{7}{20} \times \frac{5}{19} = \frac{35}{380} \]

Scenario 2: Selecting a banana first, then an apple

  1. Probability of selecting a banana first: \[ P(\text{Banana first}) = \frac{5}{20} = \frac{1}{4} \]
  2. After selecting a banana, we have:
    • 4 bananas (since we have taken 1 banana)
    • 8 mangoes
    • 7 apples

Total fruits left = 20 - 1 = 19

  1. Probability of selecting an apple next: \[ P(\text{Apple second | Banana first}) = \frac{7}{19} \]

So, the probability of selecting a banana first and then an apple is: \[ P(\text{Banana, then Apple}) = P(\text{Banana first}) \times P(\text{Apple second | Banana first}) = \frac{1}{4} \times \frac{7}{19} = \frac{7}{76} \]

Total Probability

\[ P(\text{Apple and Banana}) = P(\text{Apple, then Banana}) + P(\text{Banana, then Apple}) = \frac{35}{380} + \frac{7}{76} \] First, convert \(\frac{7}{76}\) to have the same denominator as \(\frac{35}{380}\): \[ \frac{7}{76} = \frac{35}{380} \] Now, let's simplify: \[ \frac{35}{380} + \frac{35}{380} = \frac{70}{380} = \frac{7}{38} \]

Final Answer

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

Since none of the provided options matches this probability, the closest option is \( \frac{7}{76} \) from the computation for selecting an apple followed by a banana.