Question

In a certain Algebra 2 class of 25 students, 13 of them play basketball and 11 of them play baseball. There are 4 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:

$\frac{4}{5}$

Step-by-step explanation:

Given: There are 2 classes of 25 students.

            13 play basketball

            11 play baseball.

            4 play neither of sports.

Lets assume basketball as "a" and baseball as "b".

We know, probablity dependent formula; P(a∪b)= P(a)+P(b)-p(a∩b)

As given total number of student is 25

Now, subtituting the values in the formula.

⇒P(a∪b)= $\frac{13}{25} +\frac{11}{25} -\frac{4}{25}$

taking LCD as 25 to solve.

⇒P(a∪b)= $\frac{13\times 1+11\times 1-4\times 1}{25} = \frac{20}{25}$

∴ P(a∪b)= $\frac{4}{5}$

Hence, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{4}{5}$.