Probability Calculation: Sequential Fruit Removal

This problem involves calculating the probability of two dependent events occurring in sequence.

Initial Setup

• Total fruits in the box: 3 apples + 2 oranges = 5 fruits.
• We are removing two fruits randomly in succession (without replacement).
• We need the probability that the first fruit is an apple AND the second fruit is an orange.

Step 1: Probability of the First Fruit Being an Apple

The probability of the first fruit selected being an apple is the number of apples divided by the total number of fruits.

Let $P(\text{1st is Apple})$ denote this probability.

$ P(\text{1st is Apple}) = \frac{\text{Number of Apples}}{\text{Total Fruits}} = \frac{3}{5} $

Step 2: Probability of the Second Fruit Being an Orange (Given the First was an Apple)

After removing one apple, the box composition changes:

• Remaining apples: 3 - 1 = 2
• Remaining oranges: 2
• Total remaining fruits: 5 - 1 = 4

The probability of the second fruit selected being an orange, given that the first was an apple, is the number of remaining oranges divided by the total remaining fruits.

Let $P(\text{2nd is Orange} | \text{1st is Apple})$ denote this conditional probability.

$ P(\text{2nd is Orange} | \text{1st is Apple}) = \frac{\text{Number of Oranges}}{\text{Total Remaining Fruits}} = \frac{2}{4} = \frac{1}{2} $

Step 3: Combined Probability

The probability of both events happening in sequence is the product of their individual probabilities (using the rule $P(A \text{ and } B) = P(A) \times P(B|A)$).

$ P(\text{1st is Apple AND 2nd is Orange}) = P(\text{1st is Apple}) \times P(\text{2nd is Orange} | \text{1st is Apple}) $

$ P = \frac{3}{5} \times \frac{2}{4} $

$ P = \frac{6}{20} $

Simplifying the fraction:

$ P = \frac{3}{10} $

Conclusion

The probability that the first fruit removed is an apple and the second is an orange is $\frac{3}{10}$.