There are 3 red balls, 4 blue balls and 5 white balls. 2 balls are chosen randomly. Find probability that 1 is red and the other is white.
5/22
5/23
7/22
4/9
None of These

To solve this problem, let's first calculate the total number of ways to choose 2 balls from the total set.

### Step 1: Calculate Total Number of Outcomes
There are \(3 + 4 + 5 = 12\) balls in total. The number of ways to choose 2 balls from 12 is given by the combination formula:

\[
\text{Total outcomes} = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66
\]

### Step 2: Calculate Favorable Outcomes
We want to find the probability that 1 ball is red and the other is white. 

- Number of ways to choose 1 red ball out of 3:

\[
\text{Red ball choices} = \binom{3}{1} = 3
\]

- Number of ways to choose 1 white ball out of 5:

\[
\text{White ball choices} = \binom{5}{1} = 5
\]

Thus, the number of favorable outcomes is:

\[
\text{Favorable outcomes} = 3 \times 5 = 15
\]

### Step 3: Calculate the Probability
The probability of selecting 1 red and 1 white ball is given by:

\[
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{15}{66} = \frac{5}{22}
\]

The correct answer is \( \frac{5}{22} \).

### Answer: \( \boxed{\frac{5}{22}} \)

Do you need any more details or have any questions? Here are 5 related questions you might find interesting:

1. What is the probability of selecting 2 balls of the same color?
2. If 2 blue balls are chosen randomly, what is the probability?
3. How does the probability change if there were 6 white balls instead of 5?
4. What is the probability that neither of the selected balls is blue?
5. How would the probability change if we chose 3 balls instead of 2?

**Tip:** When calculating probabilities, always break down the problem into smaller parts, such as finding the number of favorable outcomes and the total possible outcomes. This makes complex problems easier to handle.

Calculate the probability of selecting 1 red ball and 1 white ball from a set of 3 red, 4 blue, and 5 white balls. Step-by-step solution using combinatorics and probability concepts. Suitable for Grades 9-12.

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