STAT 3025

A Brief Course in Mathematical Statistics

Random Experiments: Experiments for which the outcome cannot be predicted with certainty.
Outcome Space: Collection of every possible outcome can be described and perhaps listed. Denoted by O.

EX: Roll one die. Let the outcome be the number of spots on the side that is “up.”
Therefore the outcome space is O = {1,2,3,4,5,6}

*0 with a / through it denotes a null or empty set.
*A C B means A is a subset of B
*A U B is the union of A and B
*A n B is the intersection of A and B
*A’ is the complement of A (i.e., all elements in O that are not in A).

Definition 1.1-1:
(a) P(A) greater equal to 0
(b) P(O) = 1
(c) If A1,A2,A3,… are events and Ai n Aj = 0/, i =/ j

Theorem 1.1-1:

For each event A,
P(A) = 1 – P(A’)

We have:
o = A U A’ and A n A’ = 0/.

Thus, from properties (b) and (c), it follows that:

1 = P(A) = P(A’)

Hence: P(A) = 1 – P(A’)

Theorem 1.1-2:
P(0/) = 0.
In theorem 1.1-1, take A = 0/ so that A’ = O. Thus
P(0/) = 1 – P(O) = 1-1=0

Homework #1

1.1-2)

A coin is tossed four times, and the sequence of heads and tails is observed.
(a) List each of the 16 sequences in the outcome space O.

So we can let h = Heads, and t = Tails
O = {hhhh,hhht,hhth,hthh,htth,httt,tttt,ttth,ttht,thth,thtt,thhh,thht,htht,tthh,tthh}

=> In all that is 16 possible outcomes

(b) Rules:
A = {at least 3 heads}, B = {at most 2 heads}, C = {heads on the third toss}, D = {1 head and 3 tails}.
Probability set assigns 1/16 to each outcome in the outcome space…

(i) P(A) = 5/16 [O = {hhht,hhth,hthh,thhhh,hhhh}]
(ii) P(A n B) = 0 [From the outcome of both, we don't see any that intersect]
(iii) P(B) = 10/16 => 5/8 [O = {hhtt,htth,htht,tthh,thth,thht,httt,ttth,thtt,ttht}]
(iv) P(A u C) = 4/16 => 1/4 [From A and C solutions, we see that there are 4 outcomes that intersect]
(v) P(D) = 4/16 > 1/4 [O = {httt,ttth,ttht,thtt}]
(vi) P(A U C) = 9/16 [From A and C we can see which ones include each other and ones that differ from each set]
(vii) P(B n D) = 4/16 => 1/4 [From B and D we can see where their are similar outcomes for each]

1.1-5)
If O = A U B, P(A) = 0.7 and P(B) = 0.9, find P(A n B).

"The probability of the occurrence of an event can be expressed as a fraction or a decimal from 0 to 1. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1.*"

P(A U B) = 1
=> P(A U B) = P(A) + P(B) – P(A n B)
=> P(A n B) = P(A) + P(B) – P(A U B) = 0.7 + 0.9 – 1 = 0.6

1.1-6b)
If P(A) = 0.4, P(B) = 0.5, and P(A U B) = 0.7, find (b) P(A’ U B’).

P(A U B) = P(A) + P(B) – P(A n B)
=> P(A n B) = P(A) + P(B) – P(A U B)
=> 0.4 + 0.5 – 0.7
=> 0.2

*P(A’ U B’) = 1 – P(A n B)
=> 1 – 0.2 = 0.8

*A’ U B’ includes points except the points in A n B. A’ includes everything except from A, but also includes points from B that aren’t from (A n B), and vice versa for B’. Therefore, we can say that (A’ U B’) includes all points except those in (A n B).

Other Info:
P(A’) = 1 – P(A)
P(B’) = 1 – P(B)

1.1-7)
“A typical roulette wheel used in a casino has 38 slots that are numbered 1,2,3,….,36,0,00, respectively. The 0 and 00 slots are colored green. Half of the remaining slots are red and half are black. Also half of the integers between 1 and 36 inclusive are odd, half are even, and 0 and 00 are defined to be neither odd nor eve. A ball is rolled around the wheel and ends up in one of the slots; we assume each slot has equal probability of 1/38 and we are interested in the number of the slot in which the ball falls.

(a) Define the outcome space O.
[O = {1,2,3,...,36,0,00}] – Since any of the 38 slots are a possible outcome

(b)Let A = {0,00}. Give the value of P(A).
P(A) = 2/38 => 1/19 [They gave us 2 outcomes of the 38]

(c) Let B = {14,15,17,18}. Give the value of P(B).
P(b) = 4/36 => 1/9 [They gave us 4 outcomes of the 38]

(d) Let D = {x:x is odd}. Give the value of P(D).
P(D) = 18/36 => 9/19 [From the problem, they said {0,00} aren’t considered odd/even. So we only have 36 possible outcomes. Another rule is that we want odd values. Half of 36 is even, and the other is odd.

Homework #2