in a certain algebra 2 class of 24 students, 14 of them play basketball and 5 of them play baseball. there…

Explanation:

Step1: Use the principle of inclusion - exclusion

Let $A$ be the set of basketball - playing students and $B$ be the set of baseball - playing students. The total number of students $n(T)=24$, $n(A) = 14$, $n(B)=5$, and the number of students who play neither $n(\overline{A\cup B}) = 7$. First, find $n(A\cup B)$. We know that $n(T)=n(A\cup B)+n(\overline{A\cup B})$. So, $n(A\cup B)=n(T)-n(\overline{A\cup B})=24 - 7=17$.

Step2: Apply the inclusion - exclusion formula

The inclusion - exclusion formula is $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. We know $n(A\cup B) = 17$, $n(A)=14$, and $n(B)=5$. Substitute these values into the formula: $17=14 + 5-n(A\cap B)$. Solve for $n(A\cap B)$: $n(A\cap B)=14 + 5-17=2$.

Step3: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is $P=\frac{n(A\cap B)}{n(T)}$. Since $n(A\cap B)=2$ and $n(T)=24$, then $P=\frac{2}{24}=\frac{1}{12}$.

Answer:

$\frac{1}{12}$