From an urn containing no red balls, nB black balls, and nW white balls (all identical except for color) 
3 balls are drawn at random. Calculate the following probabilities:
(i) All 3 balls are red. (ii) At least one ball is red.
(iii) One ball is red, 1 is black, and 1 is white. Do this when the balls are drawn:
(a) Successively and with replacement (b) Without replacement.

Given the problem, we need to calculate the probabilities under two different scenarios: (a) when balls are drawn successively with replacement, and (b) when balls are drawn without replacement.

### Scenario Overview:
- **Number of Red Balls**: 0 (since the urn contains no red balls)
- **Number of Black Balls**: \( n_B \)
- **Number of White Balls**: \( n_W \)
- **Total Number of Balls**: \( n_B + n_W \)

### **(i) Probability that All 3 Balls are Red**

#### (a) **With Replacement:**

Since there are no red balls in the urn, the probability of drawing a red ball in any draw is \(0\).

\[
P(\text{All 3 balls are red}) = 0 \times 0 \times 0 = 0
\]

#### (b) **Without Replacement:**

Similarly, without replacement, since there are no red balls in the urn, it is impossible to draw even one red ball.

\[
P(\text{All 3 balls are red}) = 0
\]

### **(ii) Probability that At Least One Ball is Red**

#### (a) **With Replacement:**

Since there are no red balls, it's impossible to draw a red ball.

\[
P(\text{At least one ball is red}) = 0
\]

#### (b) **Without Replacement:**

Similarly, without replacement, it’s impossible to draw a red ball.

\[
P(\text{At least one ball is red}) = 0
\]

### **(iii) Probability that One Ball is Red, One is Black, and One is White**

#### (a) **With Replacement:**

Since there are no red balls, the probability of getting one red ball is 0. 

\[
P(\text{1 Red, 1 Black, 1 White}) = 0
\]

#### (b) **Without Replacement:**

Similarly, without replacement, it is impossible to draw one red, one black, and one white since there are no red balls.

\[
P(\text{1 Red, 1 Black, 1 White}) = 0
\]

### Summary of Results:

For both scenarios:
1. **(i) Probability that all 3 balls are red:** 0
2. **(ii) Probability that at least one ball is red:** 0
3. **(iii) Probability that one ball is red, one is black, and one is white:** 0

Would you like more details on the steps or any further explanation?

Here are 5 related questions you might be interested in:

1. What if the urn had 1 red ball, and how would that affect the probabilities?
2. How would the probabilities change if you increased the number of balls in the urn?
3. Can you explain the concept of conditional probability in a similar problem?
4. What are the probabilities if the drawing is done without replacement but with a different number of balls?
5. How would these probabilities change if there were two urns with different compositions?

**Tip:** When dealing with problems involving drawing without replacement, it’s important to keep track of how the total number of available choices decreases with each draw.

Explore the probabilities of drawing balls from an urn with no red balls, nB black balls, and nW white balls. Calculate probabilities such as all balls being red, at least one ball being red, and one red, one black, and one white ball under different scenarios: with replacement and without replacement.

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