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: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.