ST104b - Statistics 2 - Solutions to all learning activities and sample examination questions from the guide

ST104b Statistics 2
Solutions to Learning activities from Chapter 2

1. Because there is no need to list the elementary outcome 1 twice. It is much clearer to write
S = {1, 2}.


2. The possible events are {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} (the sample space S) and ∅.


3. We have:

A ∩ S = A, A ∩ ∅ = ∅,
A ∪ S = S, A ∪ ∅ = A.



4. S has 4 elementary outcomes that are equally likely, so each elementary outcome has probability
1/4.

P (A ∩ B) P (c) 1/4
P (A | B) = = = = 1/2.
P (B) P (c, d) 1/4 + 1/4


P (B ∩ A) P (c) 1/4
P (B | A) = = = = 1/3.
P (A) P (a, b, c) 1/4 + 1/4 + 1/4



5. Note carefully here that we just assume equally likely elementary outcomes, so that each has
probability 1/4, and the independence follows.
The event ‘Heads on the first toss’ is A = {HH, HT } and has probability 1/2, for it is specified
by two elementary outcomes. The event ‘Heads on the second toss’ is B = {HH, T H} and has
probability 1/2. The event ‘Heads on both the first and the second toss’ is A ∩ B = {HH} and
has probability 1/4. So the multiplication property P (A ∩ B) = 1/4 = 1/2 × 1/2 = P (A) P (B)
is satisfied, and the two events are independent.


6. It is important to get the logical flow in the right direction here. We are told that A and B are
disjoint events, that is:
A ∩ B = ∅.
So:
P (A ∩ B) = 0.
We are also told that A and B are independent, that is:

P (A ∩ B) = P (A) P (B).

It follows that:
0 = P (A) P (B)
and so either P (A) = 0 or P (B) = 0.