Determine the intersection and union of two subsets of a sample space, Practice
Set C
Name:
Date:
1. Four students, A, B, C, and D are standing in a line. How likely is it that student A is
standing anywhere to the left of student B and B is standing anywhere to the left of student
C?
2. A couple plans to have four children. How likely is it that they will have more
boys than girls or more girls than boys?
3.
A red and a green die are rolled. How likely is it that at least one die is a 3 and
at most one die is a 3?
Determine the intersection and union of two subsets of a sample space, Practice
Set C
Answer Key
1. Four students, A, B, C, and D are standing in a line. How likely is it that student A is
standing anywhere to the left of student B and B is standing anywhere to the left of student
C? According to the problem, the A, B, and C need to be in alphabetical order from left to
right. D can be anywhere, so there are 4 possibilities: (A,B,C,D), (A,B,D,C),(A,D,B,C),(D,A,B,C).
There are 24 (4 x 3 x 2 x 1) total ways the students could be arranged, but only 4 of these
meet the conditions. [Note: Some students may express their answer in terms of probability
=6/24 = ¼]
2. A couple plans to have four children. How likely is it that they will have more
boys than girls or more girls than boys? By birth order, there are 5 ways in which
the couple could have more boys than girls: (B,B,B,B), (B,B,B,G), (B,B,G,B),
(B,G,B,B),(G,B,B,B) and 5 ways in which the couple could have more girls than boys:
(G,G,G,G), (G,G,G,B), (G,G,B,G), (G,B,G,G),(B,G,G,G). These don’t overlap, so there are a
total of 10 ways. There are 16 = (2 x 2 x 2 x 2) total possibilities, so it is more likely
that a couple will have more of one gender than the other - rather than two of
each.[Note: some students may express the answer in terms of probability = 10/16
=⅝]
3.
A red and a green die are rolled. How likely is it that at least one die is a 3 and
at most one die is a 3?
There are eleven ways to have at least one 3:
(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6)
There are 35 (36 - 1) ways to have at most one 3. The only option that needs to be
excluded is (3,3)
The intersection of these two sets are just those results which have exactly one 3.
(1,3),(2,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6) and there are ten of
these. So ten of the 36 possible outcomes for two dice have exactly
one 3. So this outcome is relatively likely. [Note: Some students may
express their answer in terms of probability = 10/36 = 5/18]