Step 1: The probability that the student studied for at most 2 hours is given by: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). \] Step 2: Substitute the given values: \[ P(X \leq 2) = 0.1 + k(1) + k(2). \] Step 3: Substitute \( k = 0.15 \): \[ P(X \leq 2) = 0.1 + 0.15(1) + 0.15(2). \] Step 4: Compute the value: \[ P(X \leq 2) = 0.1 + 0.15 + 0.3 = 0.55. \] Thus, the probability that the student studied for at most 2 hours is 0.55.
Match the LIST-I with LIST-II
| LIST-I (Expressions) | LIST-II (Values) | ||
|---|---|---|---|
| A. | \( i^{49} \) | I. | 1 |
| B. | \( i^{38} \) | II. | \(-i\) |
| C. | \( i^{103} \) | III. | \(i\) |
| D. | \( i^{92} \) | IV. | \(-1\) |
Choose the correct answer from the options given below:

Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below:
Based on the above data, predict the order of the reaction and write the expression for the rate law.