at a high school, students can choose between three art electives, four history electives, and five computer…

at a high school, students can choose between three art electives, four history electives, and five computer electives. each student can choose two electives. which expression represents the probability that a student chooses an art elective and a history elective?\n$\frac{_{7}c_{2}}{_{12}c_{2}}$\n$\frac{_{7}p_{2}}{_{12}p_{2}}$\n$\frac{(_{3}c_{1})(_{4}c_{1})}{_{12}c_{2}}$\n$\frac{(_{3}p_{1})(_{4}p_{1})}{_{12}p_{2}}$\n

at a high school, students can choose between three art electives, four history electives, and five computer electives. each student can choose two electives. which expression represents the probability that a student chooses an art elective and a history elective?\n$\frac{_{7}c_{2}}{_{12}c_{2}}$\n$\frac{_{7}p_{2}}{_{12}p_{2}}$\n$\frac{(_{3}c_{1})(_{4}c_{1})}{_{12}c_{2}}$\n$\frac{(_{3}p_{1})(_{4}p_{1})}{_{12}p_{2}}$\n

Answer

Explanation:

Step1: Calculate total number of electives

There are 3 art + 4 history + 5 computer = 12 electives. The number of ways to choose 2 electives out of 12 is given by the combination formula ({n}C{r}=\frac{n!}{r!(n - r)!}), so ({12}C{2}=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}).

Step2: Calculate number of ways to choose an art and a history elective

The number of ways to choose 1 art elective out of 3 is ({3}C{1}=\frac{3!}{1!(3 - 1)!}=3), and the number of ways to choose 1 history elective out of 4 is ({4}C{1}=\frac{4!}{1!(4 - 1)!}=4). By the multiplication - principle, the number of ways to choose 1 art and 1 history elective is ({3}C{1}\times_{4}C_{1}=3\times4).

Step3: Calculate the probability

The probability (P) of choosing an art and a history elective is the number of favorable outcomes divided by the number of total outcomes. So (P = \frac{({3}C{1})({4}C{1})}{{12}C{2}}).

Answer:

(\frac{({3}C{1})({4}C{1})}{{12}C{2}})