Question

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 24 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

Answer 1

Answer:

$\frac{5}{29}$

Step-by-step explanation:

Let $n(A)$ represent students playing basketball, $n(B)$ represent students playing baseball.

Then, $n(A)=7$, $n(B)=24$

Let $n(S)$ be the total number of students. So, $n(S)=29$.

Now,

$P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}$

$P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}$

3 students play neither of the sport. So, students playing either of the two sports is given as:

$n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26$

∴ $P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}$

From the probability addition theorem,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Where, $P(A\cap B)$ is the probability that a student chosen randomly from the class plays both basketball and baseball.

Plug in all the values and solve for $P(A\cap B)$ . This gives,

$\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}$

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{5}{29}$