Question

In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?

Answer 1

Probability that a student chosen randomly from the class plays basketball or baseball is
`(23)/(30)` or 0.76

Explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

`P(A \cup B)=P(A)+P(B)-P(A \cap B)`---------------(1)

where

P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) =
`\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}`

P(A) =
`(19)/(30)`--------------------------(2)

Finding P(B)

P(B) =
`\frac{\text { Number of students playing baseball }}{\text{Total number of students}}`

P(B) =
`(12)/(30)`---------------------------(3)

Finding
`P(A \cap B)`

P(A) =
`\frac{\text { Number of students playing both games }}{\text{Total number of students}}`

P(A) =
`(8)/(30)`-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

`P(A \cup B)= (19)/(30) + (12)/(30) -(8)/(30)`

`P(A \cup B)= (31)/(30) -(8)/(30)`

`P(A \cup B)= (23)/(30)`