Let's determine the probability that a student chooses one art elective and one history elective out of the total number of choices.
We should first find out the number of ways a student can choose one art elective, then one history elective, and finally, the total ways to choose any two electives out of all available electives.
Step-by-step solution:
1. Total electives available:
Students can choose from:
- 3 art electives
- 4 history electives
- 5 computer electives
Thus, the total number of electives available is [tex]\( 3 + 4 + 5 = 12 \)[/tex].
2. Number of ways to choose 1 art elective:
The number of ways to choose one art elective out of three is represented by the combination formula [tex]\({}_n C_r\)[/tex]:
[tex]\[
{ }_3 C_1
\][/tex]
3. Number of ways to choose 1 history elective:
The number of ways to choose one history elective out of four is represented by the combination formula:
[tex]\[
{ }_4 C_1
\][/tex]
4. Number of ways to choose any 2 electives out of 12 total electives:
The number of ways to choose 2 electives from 12 is given by the combination formula:
[tex]\[
{ }_{12} C_2
\][/tex]
5. Probability calculation:
To find the probability that a student chooses one art elective and one history elective, we need to calculate the ratio of the number of favorable outcomes to the total number of possible outcomes:
[tex]\[
\text{Probability} = \frac{{ }_3 C_1 \times { }_4 C_1}{{ }_{12} C_2}
\][/tex]
Thus, the expression representing the probability that a student chooses one art elective and one history elective is:
[tex]\[
\boxed{\frac{\left.\left({ }_3 C_1\right) C_1\right)}{{ }_{12} C_2}}
\][/tex]
