in a certain algebra 2 class of 30 students, 11 of them play basketball and 9 of them play baseball. there…

Explanation:

Step1: Find number of students who play at least one sport

Let $n(A\cup B)$ be the number of students who play at least one sport. Total students $N = 30$ and number of students who play neither sport is $15$. So $n(A\cup B)=30 - 15=15$.

Step2: Use the inclusion - exclusion principle

Let $n(A)$ be the number of basketball - players ($n(A)=11$) and $n(B)$ be the number of baseball - players ($n(B)=9$). By the inclusion - exclusion principle $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Substitute the known values: $15 = 11+9 - n(A\cap B)$.

Step3: Solve for $n(A\cap B)$

Rearrange the equation from Step 2: $n(A\cap B)=11 + 9-15=5$.

Step4: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is $P=\frac{n(A\cap B)}{N}$. Substitute $n(A\cap B) = 5$ and $N = 30$: $P=\frac{5}{30}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$