Answer: a) 83, b) 28, c) 14, d) 28.
Step-by-step explanation:
Since we have given that
n(B) = 69
n(Br)=90
n(C)=59
n(B∩Br)=28
n(B∩C)=20
n(Br∩C)=24
n(B∩Br∩C)=10
a) How many of the 269 college students do not like any of these three vegetables?
n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C)
n(B∪Br∪C)=[tex]69+90+59-28-20-24+10=156[/tex]
So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83
b) How many like broccoli only?
n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C))
n(only Br)=[tex]90-(28+24+10)=28[/tex]
c) How many like broccoli AND cauliflower but not Brussels sprouts?
n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C)
n(Br∩C-B)=[tex]24-10=14[/tex]
d) How many like neither Brussels sprouts nor cauliflower?
n(B'∪C')=n(only Br)= 28
Hence, a) 83, b) 28, c) 14, d) 28.