Question

In a certain Algebra 2 class of 28 students, 5 of them play basketball and 21 of them play baseball. There are 5 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?Does 'neither' mean 'not basketball AND not baseball'? Or 'not basketball OR not baseball'?

Answer 1

Answer:3/28; Neither means NOT basketball AND not baseball

Step-by-step explanation:

a) Using set notation to answer the question,

Let U be universal set(total number of student)

B be those that play basketball

C be those that play baseball

BUC be those that plays either of the games

(BUC)' will them be those that play neither

BnC be those that plays both games

(U) = 28

n(B) = 5

n(C) = 21

Before we get the probability that a student chosen randomly from the class plays both basketball and baseball, we need to get BnC first

Using the formula

n(BUC)=n(B) + n(C) -n(BnC)...(1) and;

n(U)= n(BUC)+ n(BUC)'...(2)

n(BUC)= n(U) - n(BUC)'

n(BUC) = 28-5 = 23

Therefore using eqn 1,

23= 5+21-n(BnC)

23=26-n(BnC)

n(BnC)= 3

P{n(BnC)} = n(BnC)/n(U)= 3/28

b) Neither means NOT basketball AND not baseball i.e those that played NONE of the games.