A box contains 3 white balls and 2 red balls (total 5). Two draws are made without replacement. What is the probability that the second ball drawn is red (no information about the first draw is used)?

Introduction / Context:
When two balls are drawn without replacement from a small urn, the colour of the second ball can be analysed either conditionally on the first draw or by symmetry arguments. The question explicitly asks for the unconditional probability that the second ball is red, i.e., we do not observe the first draw.

Given Data / Assumptions:

• Total balls = 5 (3 white, 2 red).
• Two draws are made without replacement.
• No information about the first draw is used; draws are from a well-mixed box.

Concept / Approach:
Key fact: In simple random sampling without replacement, each position in the draw sequence is equally likely to be any given ball. Therefore, the marginal probability that the kth draw is red equals the overall red proportion in the urn: red_count / total_count. Alternatively, use the law of total probability conditioning on the first draw.

Step-by-Step Solution:
Red proportion initially = 2/5.Therefore, P(second is red) = 2/5 by symmetry of positions.Check by conditioning: P(2nd R) = P(1st W)*P(2nd R|1st W) + P(1st R)*P(2nd R|1st R) = (3/5)*(2/4) + (2/5)*(1/4) = (6/20) + (2/20) = 8/20 = 2/5.

Verification / Alternative check:
The symmetry argument applies for any position: at each draw, without conditioning on previous outcomes, the chance of red remains the initial fraction.

Why Other Options Are Wrong:
3/5 and 21/25 overstate the likelihood; 8/25 confuses a conditional product with the unconditional calculation.

Common Pitfalls:
Confusing the unconditional probability with probabilities conditional on the first draw being white or red.

Final Answer:
2/5