There are 5 balls, 2 marked "Yes" and 3 unmarked. Five people draw the balls one by one without replacement. The question asks if the probability that each person draws a "Yes" ball is the same.

1. Problem Description

There are 5 balls, 2 marked "Yes" and 3 unmarked. Five people draw the balls one by one without replacement. The question asks if the probability that each person draws a "Yes" ball is the same.

2. Solution Steps

Let $P(i)$ be the probability that the $i$-th person draws a "Yes" ball.

For the first person, the probability of drawing a "Yes" ball is:

$P(1) = \frac{\text{Number of "Yes" balls}}{\text{Total number of balls}} = \frac{2}{5}$

For the second person, we consider two cases:

Case 1: The first person drew a "Yes" ball. Then there is 1 "Yes" ball and 3 unmarked balls remaining, for a total of 4 balls. The probability of the second person drawing a "Yes" ball is $\frac{1}{4}$. The probability of this case happening is $\frac{2}{5}$.

Case 2: The first person drew an unmarked ball. Then there are 2 "Yes" balls and 2 unmarked balls remaining, for a total of 4 balls. The probability of the second person drawing a "Yes" ball is $\frac{2}{4} = \frac{1}{2}$. The probability of this case happening is $\frac{3}{5}$.

Thus, the probability of the second person drawing a "Yes" ball is:

$P(2) = P(\text{1st person drew "Yes"}) \cdot P(\text{2nd person draws "Yes"} | \text{1st person drew "Yes"}) + P(\text{1st person drew unmarked}) \cdot P(\text{2nd person draws "Yes"} | \text{1st person drew unmarked})$

$P(2) = \frac{2}{5} \cdot \frac{1}{4} + \frac{3}{5} \cdot \frac{2}{4} = \frac{2}{20} + \frac{6}{20} = \frac{8}{20} = \frac{2}{5}$

Now, let's check for the third person. We will consider two cases to see if the 3rd person draws a "Yes" ball:

Case 1: Two "Yes" were drawn in the first two draws. The probability of this is $\frac{2}{5} * \frac{1}{4} = \frac{2}{20}$

The 3rd person's probability to draw "Yes" is 0 since there are no "Yes" balls left, $\frac{2}{20}*0 = 0$

Case 2: One "Yes" and one unmarked were drawn in the first two draws. The probability of this is $\frac{2}{5} * \frac{3}{4} + \frac{3}{5} * \frac{2}{4} = \frac{6}{20} + \frac{6}{20} = \frac{12}{20}$

The 3rd person's probability to draw "Yes" is $\frac{1}{3}$, so the probability of this case is $\frac{12}{20} * \frac{1}{3} = \frac{12}{60}$

Case 3: Two unmarked were drawn in the first two draws. The probability of this is $\frac{3}{5} * \frac{2}{4} = \frac{6}{20}$

The 3rd person's probability to draw "Yes" is $\frac{2}{3}$, so the probability of this case is $\frac{6}{20} * \frac{2}{3} = \frac{12}{60}$

Adding the probabilities, we get $\frac{12}{60} + \frac{12}{60} = \frac{24}{60} = \frac{2}{5}$

In general, the probability of drawing a "Yes" ball is the same for each person, which is $\frac{2}{5}$.

3. Final Answer

Yes, the probability that each person draws a "Yes" ball is the same.