A box holds 3 mangoes and 3 apples (6 fruits total). Two fruits are chosen uniformly at random without replacement. What is the probability that one fruit is a mango and the other is an apple?

Introduction / Context:
This is a classic two-draw mixture question. With equal counts of two fruit types, we need the probability of selecting a mixed pair (one of each) when drawing two without replacement.

Given Data / Assumptions:

• Mangoes = 3, Apples = 3.
• Total fruits = 6.
• Two draws without replacement.

Concept / Approach:
Count favourable unordered pairs and divide by the total number of unordered pairs. Alternatively, compute sequentially: P(mango then apple) + P(apple then mango).

Step-by-Step Solution (combinatorial):
Total unordered pairs = C(6,2) = 15.Favourable pairs (one of each) = 3 * 3 = 9 (choose 1 mango and 1 apple).Probability = 9/15 = 3/5.

Verification / Alternative check (sequential):
P(M then A) = (3/6)*(3/5) = 3/10.P(A then M) = (3/6)*(3/5) = 3/10.Total = 3/10 + 3/10 = 3/5.

Why Other Options Are Wrong:
2/3 and 1/3 are typical but incorrect guesses; 1/5 is far too small.

Common Pitfalls:
Forgetting that order does not matter for the event “one of each” but it matters in sequential computation (hence we add both orders).

Final Answer:
3/5