Q.An experiment consists of rolling a die until a 2 appears.

Concept: Set Operations — Counting sequences with a fixed termination condition.

1. For the 2 to appear exactly on the $k$th roll, the first $k-1$ rolls must each be not 2 (5 choices each: 1,3,4,5,6), and the $k$th roll must be 2 (1 choice). Number of such sequences = $5^{k-1} \times 1 = 5^{k-1}$.
2. The event “2 appears not later than the $k$th roll” means the 2 appears on the 1st, or 2nd, …, or $k$th roll. These are mutually exclusive cases. …

Why this works

When you roll a fair die, each roll has 6 equally likely outcomes: $\{1,2,3,4,5,6\}$. The experiment stops the moment a 2 appears. So the sample space consists of all finite sequences of rolls where the last roll is a 2, and none of the earlier rolls are 2.

This is a classic "waiting time" problem. The structure is simple: before the 2 appears, every roll must be one of the other 5 numbers. Once the 2 shows up, the sequence ends.

Step-by-step solution

1. Understanding the event for part (i)

We want the number of sequences where the 2 appears exactly on the $k$th roll.

That means:

• Rolls $1, 2, \dots, k-1$: each must be not 2 — so each has 5 possible outcomes ($1,3,4,5,6$).
• Roll $k$: must be exactly 2 — only 1 outcome.

Since the choices for each roll are independent, the total number of such sequences is:

2. Answer for part (i)

The number of elements in the sample space where the 2 appears on the $k$th roll is:

3. Understanding the event for part (ii)

Now we want the 2 to appear not later than the $k$th roll. That means it could appear on the 1st, 2nd, 3rd, …, or $k$th roll.

These are mutually exclusive events — the 2 can't appear on two different rolls in the same sequence. So we simply add the counts for each possible success roll.

From part (i), the number of sequences where the 2 appears on the $r$th roll is $5^{r-1}$.

So the total for rolls $1$ through $k$ is:

$1 + 5 + 5^2 + \cdots + 5^{k-1}$ …