Ten cards numbered 1 to 10 are drawn twice with replacement, so the draws are independent. The probability that the second draw exceeds the first requires averaging the conditional probabilities over all possible first draws.

Probability Calculation

Each draw has 10 equally likely outcomes, yielding 100 possible ordered pairs (first, second). The favorable cases occur when second > first.

For first draw = 1, second can be 2-10 (9 cases).
For first draw = 2, second can be 3-10 (8 cases).
…
For first draw = 9, second can be 10 (1 case).
For first draw = 10, second cannot exceed (0 cases).

Total favorable outcomes: $9+8+\cdots +1=45$. Thus, probability = 45100=0.4510045=0.45.​

Introduction to Probability Second Card Greater Than First

In probability problems involving cards numbered 1-10 placed face down, drawing twice with replacement tests understanding of independent events. The key phrase “probability second card greater than first” appears in competitive exams, requiring exact calculation of favorable outcomes over total possibilities.

Step-by-Step Solution

• Total outcomes: $10\times 10=100$, since replacement keeps draws independent.

• • Favorable: Sum cases where second > first, from
    ∑_k=1⁹(10 − k) =
    ^{9 × 10}⁄₂ = 45.

• Probability: $0.45$, rounded to two decimal places as required.​

Why No Options Provided?

The query mentions “explain every option,” but this fill-in-the-blank style question (common in IIT JAM) has no multiple-choice options. The direct computation confirms 0.45 without alternatives.

Exam Tips for Similar Problems

• Recognize replacement implies uniform 110101 probability per card.

• Use symmetry: P(second > first) = P(second < first) = 0.45; P(equal) = 0.10 sums to 1.

• Practice verifies: Explicit enumeration matches formula.