Answer : Option E.
There are 3 red balls, 4 blue balls and 5 white balls. 2 balls are chosen randomly.
probability that 1 is red and the other is white = 3/12× 5/11 = 5/44

Let's clarify the approach used here and understand where the discrepancy arises.

### Step 1: Correcting the Approach

You used a method that involves directly multiplying probabilities:

1. **Probability of selecting a red ball first**: There are 3 red balls out of 12, so the probability is:
   \[
   \frac{3}{12} = \frac{1}{4}
   \]

2. **Probability of selecting a white ball second**: After removing one red ball, 11 balls remain, with 5 being white:
   \[
   \frac{5}{11}
   \]

The combined probability of selecting a red ball first and a white ball second is:
\[
\frac{1}{4} \times \frac{5}{11} = \frac{5}{44}
\]

### Step 2: Considering All Possible Orders

This calculation assumes the specific order of drawing (red first, white second). However, the balls could be drawn in any order (red first or white first), so we should consider both possible orders:

1. **Red first, white second**: \( \frac{1}{4} \times \frac{5}{11} = \frac{5}{44} \)
2. **White first, red second**: \( \frac{5}{12} \times \frac{3}{11} = \frac{15}{132} = \frac{5}{44} \)

Adding these two scenarios gives:
\[
\frac{5}{44} + \frac{5}{44} = \frac{10}{44} = \frac{5}{22}
\]

So, the probability remains \( \frac{5}{22} \), which is consistent with our earlier approach.

### Conclusion:
The correct answer remains \( \frac{5}{22} \), not \( \frac{5}{44} \). 

It seems there was a misunderstanding in applying the method correctly. The given correct answer of \( \frac{5}{44} \) is valid only for one specific order of drawing. Since the problem does not specify an order, the correct method should account for both possible orders, leading to the final probability of \( \frac{5}{22} \).

Let me know if you need any more explanations or have other questions! Here are 5 related questions to explore:

1. What is the probability of choosing 2 balls that are neither red nor white?
2. How would the probability change if we replace one red ball with a blue ball?
3. What is the probability that both selected balls are red?
4. If 3 balls are chosen, what is the probability that one of each color is selected?
5. How would the probability change if we remove one white ball before making the selection?

**Tip:** Always double-check whether the problem considers different orders (e.g., first red then white vs. first white then red). This ensures that your probability calculations account for all possible outcomes.

Understand how to calculate the probability of selecting one red and one white ball from a set of red, blue, and white balls. Learn the correct method and explore related probability concepts. Suitable for Grades 10-12.

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