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.
