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

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  $\frac{23}{30}$ or 0.76

Step-by-step 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) = $\frac{19}{30}$--------------------------(2)

Finding  P(B)

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

P(B) = $\frac{12}{30}$---------------------------(3)

Finding $P(A \cap B)$

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

P(A) = $\frac{8}{30}$-----------------------------(4)

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

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{31}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{23}{30}$