Download Vrije Universiteit Brussel First year master`s program in human

Vrije Universiteit Brussel
First year master’s program in human ecology
Probability and statistics
Ronald Buyl
1. Probability
I. Theory :
I.1 Definitions :
V : the set of possible solutions.
An event is a subset of V.
Q(V) is the set of all events defined on V.
The probability P is a function Q(V) → [0,1] satisfying:
∀ A ∈ Q(V) : 0 ≤ P(A) ≤ 1
P(∅) = 0, P(V) = 1
If A∩B = ∅ ⇒ P(A∪B) = P(A) + P(B)
If ∀ i, j with i ≠ j, Ai ∩ Aj = ∅ ⇒
P(A1∪A2∪…∪An) = P(A1) + P(A2) + … + P(An)
A And B are independent if P(A∩B) = P(A)⋅P(B)
Conditional probability:
P(A|B) = P(A∩B)/P(B)
P(B|A) = P(B∩A)/P(A)
P(A|B) ⋅ P(B)
Bayes : P(B| A) =
P(A| B) ⋅ P(B) + P(A| B) ⋅ P(B)
Further simplification can be obtained by replacing P(B) with 1 - P( B ).
I.2 Properties:
If ⊂ B ⇒ P(A) ≤ P(B)
P( A ) = 1 - P(A)
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∩B) = P(A|B)⋅P(B) = P(B|A)⋅P(A)
P(A∩B∩C) = P(A|B∩C)⋅P(B|C)⋅P(C)
II. Exercises :
1. A sample of American women was grouped by hair- and eyecolor:
Blue (X)
Green (Y)
Brown (Z)
Total
Blond (A)
1768
946
115
2829
Brown (B) Black (C)
807
189
1387
708
438
288
2632
1185
Red (D)
47
53
16
116
Total
2811
3094
857
6762
a. Determine P(A), P(X), P(A∩X) and P(A∪X)
b. Determine P( A ) and P( Y )
c. Determine the conditional probabilities P(A|X) and P(X|A)
d. Are A and Z independent ?
2. Events A, B and C are defined in Q(V). The respective results of the events A, B and
C do not overlap, i.e. the mutual intersections of the events are empty. The union of A, B
and C is equal to V. We further know that the probability of event A occurring is twice
the probability on event B and that the probability on event B is twice the probability on
event C.
Determine P(A), P(B) and P(C).
3. In a certain school, we know that 60% of all students are younger than 21 and that 50%
of the students are girls. 30% of the students are girls and younger than 21 years. What is
the probability that a random chosen student is a boy not younger than 21?
4. Consider a diagnostic test for a certain kind of cancer for which the following results
are known:
- The probability of a positive result if the patient has cancer is 96%.
- The probability of a negative result on a healthy person is 95%.
Considering a population where the probability on cancer is 1/200 and calculate:
a. the probability that a patient with a positive result has cancer.
b. the probability that a patient with a negative result nevertheless does have cancer.
c. the probability on a correct result.
d. Repeat these calculations for populations with a probability on cancer of 1/100 and
of 1/10.
III. Additional exercises :
A. We take a sample of 1000 apples. These are divided according to weight and
symptoms of a specific infection: no symptoms, mild, medium or serious. The following
results were found:
- of the 300 apples with a weight < 60 g, there are 60 without symptoms,
70 with “mild” and 80 with “medium” symptoms.
- of the 400 apples with a weight between 60 and 80 g, there are 140 without
symptoms, 120 “medium” and 80 “serious”.
- of the class with a weight > 80 g there are 80 without symptoms, 70 with
“mild”, 100 with “medium” and the rest with “serious” symptoms.
From the above information, what is the probability that an apple with weight ≥ 60 g
shows “medium” or “serious” symptoms?
B. Regarding the new alcoholtest used by the Belgium police, we have the following
information:
The probability that a drunken driver has a positive testresult is 0,95.
The probability that a sober driver has a negative testresult is 0,97.
The probability that a driver is drunk, is 0,15.
What is the probability that a driver, who’s test showed a negative result was never the
less drunk?