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. what is the approximate probability that a student chooses a computer elective and an art elective? 0.11364 0.21212 0.22727 0.42424

at a high school, students can choose between three art electives, four history electives, and five computer electives. each student can choose two electives. what is the approximate probability that a student chooses a computer elective and an art elective? 0.11364 0.21212 0.22727 0.42424

Answer

Explanation:

Step1: Calculate total number of electives

There are $3$ art + $4$ history+ $5$ computer = $12$ electives.

Step2: Calculate number of ways to choose 2 electives

Using combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, with $n = 12$ and $r=2$, we have $C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$.

Step3: Calculate number of ways to choose a computer and an art elective

Number of ways to choose 1 computer out of 5 and 1 art out of 3. Using multiplication - principle of counting, it is $C(5,1)\times C(3,1)$. Since $C(5,1)=\frac{5!}{1!(5 - 1)!}=5$ and $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$, then $C(5,1)\times C(3,1)=5\times3 = 15$.

Step4: Calculate the probability

Probability $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{15}{66}\approx0.22727$.

Answer:

0.22727