A basket contains 10 apples and 20 oranges. Of these, 3 apples and 5 oranges are defective. If two fruits are chosen at random without replacement, what is the probability that either both fruits are oranges or both fruits are non defective?

Introduction / Context:
This question combines classification into defective and non defective fruits with the probability of drawing fruits of specific types. We are asked to find the probability of the union of two events: both fruits are oranges, or both fruits are non defective. Because these events overlap, we must use the addition rule with subtraction of the intersection.

Given Data / Assumptions:

• Apples = 10, of which defective apples = 3, so non defective apples = 7.
• Oranges = 20, of which defective oranges = 5, so non defective oranges = 15.
• Total fruits = 10 + 20 = 30.
• Two fruits are drawn without replacement.
• Event A: both fruits are oranges.
• Event B: both fruits are non defective.

Concept / Approach:
We use combinations to count pairs of fruits. The total number of ways to choose 2 fruits from 30 is C(30, 2). We compute P(A), P(B), and P(A and B). The required probability P(A or B) equals P(A) + P(B) minus P(A and B). Intersection A and B means both fruits are non defective oranges.

Step-by-Step Solution:
Total pairs = C(30, 2) = 30 * 29 / 2 = 435. Event A: both oranges. Oranges = 20, so C(20, 2) = 20 * 19 / 2 = 190. Event B: both non defective. Non defective fruits = 7 + 15 = 22, so C(22, 2) = 22 * 21 / 2 = 231. A and B: both non defective oranges. Non defective oranges = 15, so C(15, 2) = 15 * 14 / 2 = 105. Favourable pairs = 190 + 231 - 105 = 316. Required probability = 316 / 435.

Verification / Alternative check:
We could also calculate probabilities directly: P(A) = (20/30) * (19/29), P(B) = (22/30) * (21/29), and P(A and B) = (15/30) * (14/29). Expanding and simplifying with a common denominator 30 * 29 gives the same result 316/435 when combined as P(A) + P(B) minus P(A and B). This confirms the combination method.

Why Other Options Are Wrong:
136/345 and 158/435 are smaller, corresponding to fewer favourable outcomes and do not match the correct counting. 17/87 is not based on any natural pair count from 435 and does not simplify to 316/435.

Common Pitfalls:
Students often forget to subtract the intersection when using P(A or B) and end up double counting cases where both fruits are non defective oranges. Another mistake is to treat the two events as mutually exclusive, which they are not. Careful classification avoids such errors.

Final Answer:
Thus, the required probability that both fruits are oranges or both are non defective is 316/435.