Transcript 1. Additive law of probability

Population Genetics Lab
Lab Instructor:
Hari Chhetri
PhD student
Department of Biology
Area of Research:
Association genetics (Populus)
Office:
Life Sciences Building, Room # 5206.
Office Hour:
appointment.
T, W, F – 11:30 AM – 12:30 PM or by
Email ID:
[email protected]
Tel. #:
304-293-6181
Probability and Population
Genetics
Population genetics is a study of probability
Sampling alleles from population each
generation
A a A A
a a
A
A A
a
Probability
Frequentist Approach
• Determine how often you
expect event A to occur
given a LONG series of trials
Bayesian Approach
• Determine the plausibility
of event A given what you
already know (prior).
Probability
Measure of chance.
P(E) = # of favorable outcome / Total # of possible outcome
It lies between 0 (impossible event) and 1 (certain event).
Ex. What is the probability of getting a head in one
toss of a balanced coin.
Total possible outcomes = 2 ( H, T)
# of Heads = 1 (H)
P(H) = 1/ 2 = 0.5 = 50 %
Sample- point method :
1.Define sample space (S): Collection of all possible outcomes of a
random expt.
Ex. S (Coin tossed twice)
Outcome
1
2
3
4
First Toss
H
H
T
T
Second Toss
H
T
H
T
Shorthand
HH
HT
TH
TT
2. Assign probabilities to all sample points
Ex. P(HH) = ¼ ; P(HT) = ¼ ; P(TH) = ¼ ; P(TT) = ¼
Sample- point method :
3.Determine event of interest and add their probabilities.
Ex. Find the probability of getting exactly one head in two tosses of a
balanced coin.
i. S (Coin tossed twice) { HH, HT, TH, TT}.
ii. P(HT) = ¼ ; P(TH) = ¼
iii. P(HT) + P (TH) = ¼ + ¼ = 2/4 = ½ .
If all sample points have equal probabilities then –
P(A) = na / N
where, na = # of points constituting event A and N= Total # of sample
points.
Sample- point method :
Example: Use the Sample Point Method to find the probability of
getting exactly two heads in three tosses of a balanced coin.
1. The sample space of this experiment is:
Outcome
1
2
3
4
5
6
7
8
Toss 1
Head
Head
Head
Tail
Tail
Tail
Head
Tail
Toss 2
Head
Head
Tail
Head
Tail
Head
Tail
Tail
Toss 3
Head
Tail
Head
Head
Head
Tail
Tail
Tail
Shorthand
HHH
HHT
HTH
THH
TTH
THT
HTT
TTT
Probabilities
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
2. Assuming that the coin is fair, each of these 8 outcomes has a probability of 1/8.
3. The probability of getting two heads is the sum of the probabilities of outcomes 2,
3, and 4 (HHT, HTH, and THH), or 1/8 + 1/8 + 1/8 = 3/8 = 0.375.
In above example, find the probability of getting at least two heads.
Solution: 1/8 + 1/8 + 1/8+ 1/8 = 1/2
Probability
Problem 1: The game of “craps” consists of rolling a pair of balanced dice (i.e., for each die
getting 1, 2, 3, 4, 5, and 6 all have equal probabilities) and adding up the resulting numbers.
A roll of “2” is commonly called “snake eyes” and causes an instant loss when rolled in the
opening round. Using the Sample-Point Method, find the exact probability of a roll of
snaked eyes. (Time : 10 minutes)
For large sample space: Use fundamental counting methods.
1.mn rule : If there are “m” elements from one group and
“n” elements from another group, then we can have “mn”
possible pairs, with one element from each group.
1
1
First die
2
3
4
5
6
mn= 6*6= 36 .
2
Second die
3
4
5
6
For large sample space : Use fundamental counting methods.
2. Permutation: Ordered set of “r” elements, chosen without
replacement, from “n” available elements.
n!
P 
(n  r )!
n
r
Remember:
0! = 1 (By definition)
n! = n*(n-1)*(n-2)*…………*2*1.
Example: How many trinucleotide sequences can be formed without
repeating a nucleotide , where ATC is different from CAT?
Solution: n = 4 ( A, T, C and G)
r=3
Prn 
4!
(4  3)!
= 24.
For large sample space : Use fundamental counting principle.
3. Combination: Unordered set of “r” elements, chosen without
replacement, from “n” available elements.
n!
C 
r!(n  r )!
n
r
Example: How many trinucleotide sequences, can be formed without
repeating a nucleotide , where ATC is same as CAT.
Solution: n = 4 ( A, T, C and G)
r=3
4!
C 
3!(4  3)!
n
r
= 4.
For large sample space : Use fundamental counting principle.
Problem 2: There are 36 computer workstations in this lab. If there are
18 students in the class, how many distinct ways could students be
arranged, with one student per workstation? ( 10 minutes).
Problem 3: A local fraternity is organizing a raffle in which 30 tickets are
to be sold  one per customer. (10 minutes).
a. What is the total number of distinct ways in which winners can be
chosen if prizes are awarded as follows:
Order of Drawing
Prize
First
$100
Second
$50
Third
$25
Fourth
10$
b. If holders of the first four tickets drawn each receive a $30 prize?
Laws of Probability
1. Additive law of probability:
P (A  B)  P(A)  P(B) - P(A  B)
Where, P(A  B)  Probabilit y of occurrence of event A or B.
P(A  B)  Probabilit y of occurrence of event A and B simaltaneo usly.
For mutually exclusive events, P(A  B)  0, then P(A  B)  P(A)  P(B)
A
B
A and B are Mutually Exclusive
Venn diagram for P (A  B)
Venn diagram for P (A  B)
Laws of Probability
1. Additive law of probability:
Example: From a pack of 52 cards, one card is drawn at random. Find the
probability that the card is “Heart” or “Ace”.
Four suits are : Spades, Diamonds, Clubs and Hearts.
Each suit has 13 cards: Ace,2,3,4,5,6,7,8,9,10,Jack, Queen and King.
There are four of each type, like 4 Aces,4 Jacks, 4 Queens, 4 Kings etc.
Solution:
P (H  A)  P(H)  P(A) - P(H  A)
13
4
1
P(H)  ; P(A)  ; P(H  A) 
52
52
52
13 4
1 16
P(H  A) 



52 52 52 52
Laws of Probability
2. Multiplicative law of probability:
P (A  B)  P(A)  P(B)
(If A and B are independent events)
\ | A)  P(B)  P(A | B)
P (A  B)  P(A)  P(B
(If A and B are dependent events)
Example: A pond consists of 50 salmon and 25 trout. Two fish
are drawn one by one. Find the probability that both fish are
Salmon.
a.)with replacement and
b.)without replacement
Solution :
Let, P(A) : Prob. that first fish drawn is Salmon.
P(B) : Prob. that second fish drawn is Salmon.
Case (a) : With replacemen t :
P(A  B)  P(A)  P(B)
50 50


75 75
4

9
Case (b) : Without replacemen t :
P(A  B)  P(A)  P(B | A)
50 49
98
49




75 74 222 111
Problem 4. An inexperienced spelunker is preparing for the exploration of a big
cave in a rural area of Mexico. He is planning to use two independent light sources
and from reading their technical specifications, he has concluded that each source is
expected to malfunction with probability of 0.01. What is the probability that:
a) At least one of his light sources malfunctions?
b) Neither of his light sources malfunction?
(Time : 15 minutes)
Problem 5. GRADUATE STUDENTS ONLY: In street craps, the opening toss wins if a 7
or 11 is rolled, and the “pass” bets will pay off. Meanwhile if 2, 3, or 12 is rolled, only
“don’t pass” bets will win.
a) Is it safest to bet “pass” or “don’t pass” on the opening roll? Show the exact
probability of each outcome.
b) If the shooter rolls 7 three times in a row, is it safest to bet “pass” or “don’t pass”
on the next roll? Defend your answer.
(Time: 5 minutes)