A bag contains $a$ white balls and $b$ red balls. $k$ individuals sequentially draw a ball from the bag. We want to calculate the probability that the $i$-th person ($i = 1, 2, ..., k$) draws a white ball under two conditions: (1) with replacement, and (2) without replacement.

1. Problem Description

A bag contains $a$ white balls and $b$ red balls. $k$ individuals sequentially draw a ball from the bag. We want to calculate the probability that the $i$-th person ($i = 1, 2, ..., k$) draws a white ball under two conditions: (1) with replacement, and (2) without replacement.

2. Solution Steps

(1) With Replacement:

When the balls are drawn with replacement, each draw is independent of the previous draws. The probability of drawing a white ball remains constant at each draw.

The total number of balls in the bag is $a + b$.

The probability of drawing a white ball is the number of white balls divided by the total number of balls.

$P(\text{white}) = \frac{a}{a + b}$

Since the balls are drawn with replacement, the probability of the $i$-th person drawing a white ball is the same as the probability of the first person drawing a white ball.

(2) Without Replacement:

Let $W_i$ be the event that the $i$-th person draws a white ball.

We want to find $P(W_i)$.

We can express $P(W_i)$ as:

$P(W_i) = P(W_i | W_1)P(W_1) + P(W_i | R_1)P(R_1)$

Here, $W_1$ denotes the event that the first person draws a white ball, and $R_1$ denotes the event that the first person draws a red ball.

We have $P(W_1) = \frac{a}{a+b}$ and $P(R_1) = \frac{b}{a+b}$.

Consider the case where $i = 2$. Then

$P(W_2) = P(W_2 | W_1)P(W_1) + P(W_2 | R_1)P(R_1) = \frac{a-1}{a+b-1}\frac{a}{a+b} + \frac{a}{a+b-1}\frac{b}{a+b} = \frac{a(a-1) + ab}{(a+b)(a+b-1)} = \frac{a(a-1+b)}{(a+b)(a+b-1)} = \frac{a(a+b-1)}{(a+b)(a+b-1)} = \frac{a}{a+b}$

Now, let's consider the general case for the $i$-th person. We will use induction.

The base case $i = 1$ is trivial, $P(W_1) = \frac{a}{a+b}$.

Assume that $P(W_{i-1}) = \frac{a}{a+b}$.

Then $P(W_i) = \frac{a}{a+b}$.

This can be shown using conditional probability and considering all possible sequences of white and red balls drawn before the $i$-th draw. However, there's a much simpler argument:

Consider all $(a+b)$ balls and label them $1, 2, \dots, a+b$.

There are $(a+b)!$ ways to order them.

We want to find the probability that the $i$-th ball is white.

There are $a$ choices for a white ball to be in the $i$-th position.

For each of these choices, there are $(a+b-1)!$ ways to arrange the other balls.

So there are $a(a+b-1)!$ ways to have a white ball in the $i$-th position.

The probability of a white ball in the $i$-th position is $\frac{a(a+b-1)!}{(a+b)!} = \frac{a}{a+b}$.

3. Final Answer

(1) With replacement: $\frac{a}{a+b}$

(2) Without replacement: $\frac{a}{a+b}$