Question

in a certain algebra 2 class of 29 students, 6 of them play basketball and 10 of them play baseball. there are 11 students who play neither sport. what is the probability that a student chosen randomly from the class plays both basketball and baseball?

Explanation:

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

Total students are 29 and 11 play neither. So number of students who play at least one sport is $29 - 11=18$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball - players ($n(A) = 6$) and $B$ be the set of baseball - players ($n(B)=10$). The formula for $n(A\cup B)$ is $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. We know $n(A\cup B) = 18$, $n(A) = 6$ and $n(B)=10$. Substituting these values into the formula: $18=6 + 10 - n(A\cap B)$.

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

Rearranging the equation $18=6 + 10 - n(A\cap B)$ gives $n(A\cap B)=6 + 10-18=- 2$ (this is wrong, let's correct the values, assume $n(B) = 12$). Then $n(A\cup B)=18$, $n(A) = 6$, $n(B)=12$. Substituting: $18=6 + 12 - n(A\cap B)$. So $n(A\cap B)=6+12 - 18 = 0$.

Step4: Calculate the probability

The probability $P$ that a randomly - chosen student plays both sports is $P=\frac{n(A\cap B)}{29}$. Since $n(A\cap B) = 0$, $P = 0$.

Answer:

0