To find the probability that a student chooses an art elective and a history elective, let's break down the problem step by step.
1. Total Number of Electives:
- There are [tex]\(3\)[/tex] art electives, [tex]\(4\)[/tex] history electives, and [tex]\(5\)[/tex] computer electives.
- Total number of electives available is [tex]\(3 + 4 + 5 = 12\)[/tex].
2. Number of Ways to Choose Two Electives:
- The total number of ways to choose 2 electives from 12 is given by the combination formula [tex]\(\binom{n}{k}\)[/tex] which represents "n choose k."
- Therefore, the number of ways to choose 2 electives out of 12 is [tex]\(\binom{12}{2}\)[/tex].
3. Number of Favorable Ways to Choose One Art and One History Elective:
- The number of ways to choose 1 art elective out of 3 is [tex]\(\binom{3}{1}\)[/tex].
- The number of ways to choose 1 history elective out of 4 is [tex]\(\binom{4}{1}\)[/tex].
- The total number of ways to choose one art and one history elective is [tex]\(\binom{3}{1} \times \(\binom{4}{1}\)[/tex].
4. Calculating the Probability:
- The probability that a student chooses an art elective and a history elective is the ratio of the number of favorable ways to the total number of ways.
- Therefore, the probability is [tex]\(\frac{\text{number of favorable ways}}{\text{total number of ways}} = \frac{\binom{3}{1} \times \(\binom{4}{1}}{\binom{12}{2}}\)[/tex].
Using the values as given:
- [tex]\(\binom{3}{1} = 3\)[/tex]
- [tex]\(\binom{4}{1} = 4\)[/tex]
- [tex]\(\binom{12}{2} = 66\)[/tex]
The probability is:
[tex]\[ \frac{3 \times 4}{66} = \frac{12}{66} = \frac{2}{11} \approx 0.1818 \][/tex]
Given the answer choices:
- The correct expression that matches our derived probability is:
[tex]\[\frac{\left( \binom{3}{1} \times \binom{4}{1} \right)}{\binom{12}{2}}\][/tex]
In terms of the provided answer choices, this matches:
[tex]\[\frac{\left.\left( { }_3 C_1 \right) C_1 C_1\right)}{12 C_2}\][/tex]
So, the correct choice for the expression is:
[tex]\[\frac{\left.\left( { }_3 C_1 \right) C_1 C_1\right)}{12 C_2}\][/tex]
