Presentation is loading. Please wait.

Presentation is loading. Please wait.

PROBABILITIES. Probabilities Basic concepts: Random experiment: All possible outcomes have to be known in advance The result of a particular experimen.

Similar presentations


Presentation on theme: "PROBABILITIES. Probabilities Basic concepts: Random experiment: All possible outcomes have to be known in advance The result of a particular experimen."— Presentation transcript:

1 PROBABILITIES

2 Probabilities Basic concepts: Random experiment: All possible outcomes have to be known in advance The result of a particular experimen tcannot be predicted (randomness). The experiment can be repeated under identical conditions Sample space Ω: Set of all possible outcomes Events A sub-set of the sample space, i.e. a set of possible otucomes Singular event: A sub-set of the sample space with one element

3 Probability model 1. Set S of possible outcomes a) Throw the coin, S = (pitch, toss) b) Throw the dice, S = (1, 2, 3, 4, 5, 6) 2. Probabilites P i a) Throw the coin, S = (pitch, toss) P(pitch) = P(toss) = 1/2 b) Throw the dice, S = (1, 2, 3, 4, 5, 6) P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

4 Probabilities Dice 1, 2, 3, 4, 5, 6 P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6 Two dice are thrown Model W2 W1 1/6 1/36

5 Probabilities 1. P(both dice show 1) = P(1)  P(1) = 1/6  1/6 = 1/ Model W2 W1 1/6

6 Probabilities 2. Event E: get a „2“ and a „3“: P(E) = 2  P(2)  P(3) = 2  1/6  1/6 = 1/18 Event F: first dice a „2“, second dice a „3“, P(F) = 1/6  1/6 = 1/ Model W2 W1 1/6

7 Probabilities 3. P(double) = P(double 1) P(double 6) = 6  1/36 = 1/ Model W2 W1 1/6

8 Probabilities 4. P(sum of both dice is 7) = 6  1/36 = 1/ Model W2 W1 1/6

9 Probabilities Two hunters try to shoot a fox. Each hunter strikes with probability 1/3. What is the probability, that the fox survives?  Each hunter does not strike with probability 2/3  P(fox survives) = P(both hunters do no strike) = 2/3 ∙ 2/3 = 4/9

10 Further probability exercises

11 PROBABILITIES Factorial, binomial coefficient and binomial distribution

12 Probabilities Factorial: A number of students is called to the black board. How many orderings are possible? 1A 1(A) n ∙(n-1) ∙  ∙2 ∙1 3 ∙2 = 6 CAB; CBA; ACB; BCA; ABC; BAC 2AB; BA 3(A, B, C) Number of possible orderings OrderingsNumber of students 2(A, B) n Factorial In R: factorial() 0! := 1 1! = 1 2! = 1∙2 = 2 3! = 1∙2 ∙3 = 6 n! = 1∙2 ∙3 ∙  ∙n

13 Probabilities Binomial coefficient Choose 6 numbers out of 49. How many outcomes are there if the ordering does NOT matter and if numbers can be selected only once? 49 · 48 · 47 · 46 · 45 · 44 choices 49 · 48 choicesFirst and 2nd number: 48 choicesSecond number: 49 choicesFirst number: 6 numbers: Is this the solution?No! – Why? 1. selection:{7; 18; 5; 43; 1; 22} 2. selection:{18; 7; 5; 43; 1; 22} Both selections are equal, since they differ only with respect to the order How many orderings are there?6! Final solution of question above:

14 Probabilities Claim: Definition: is the number of ways k things can be chosen from n things. Here, the ordering of the k things is not of interest and each element can be chosen only once. Binomial coefficient (In R: choose(n,k) )

15 Probabilities Binomial coefficient S: Phenotype:affected unaffected PKU is autosomal reccesive hereditary disease. Pedigree: m: mutant allele w: wildtype allel (= „normal“ allele) Inheritance (PKU)

16 Probabilities Binomial distribution ? S: P(child is ill)= 1/2 ∙ 1/2 =1/4 P(child is not ill)= 1- P(child is ill)= 3/4

17 Probabilities Binomial distribution S: ?? P(first child ill, 2nd child not ill) = 1/4 ∙ 3/4 =3/16 P(one child ill, one child not ill) = 2 ∙ 1/4 ∙ 3/4 =6/16

18 Probabilities S: ????? orderingshealthy ill healthy ill An experiment with two (->“binomial“) possible outcomes („success“ and „failure“) is independently repeated n times p: probability of event 1 („success“) per experiment n: number of experiments k: number of successes Binomial distribution (In R: dbinom(k,n,p) ):

19 Probabilities Binomial distribution Density function S: ????? p = 1/4 (affected)  (1-p) = 3/4 (not affected)

20 Probabilities Binomial distribution Density function

21 Probabilities Random variable x y z Probability model S Probability P Real numbers R function X X: Random variable ?

22 Probabilities Random variable: function X that assigns a real number to an event S Examples for possible random variables X, Y X: number of edges Y: red  1 brown  2 1 2

23 Probabilities Random variable S Question: P(X = 3) = ? X:number of edges (sloppy) correct: P( a  S | X(a) = 3) = Use the sloppy notation!

24 Probabilities S X:Number of edges P(X = 4) = P(X = 0) = P(X = 2) = P(X = 3) = P(X = 5) = P(X  4) = P(X  0) = P(X  2) = P(X  3) = P(X  5) = P(X < 0) = Distribution function: F(a) = P(X  a) Density 21 Distribution function 1 2

25 Probabilities Statistical measures SampleModel Expectation Example 7/204 1/43 2/50 pipi xixi MeanEmpircal varianceEmp. standard deviation Variance Standard deviation

26 Probabilities Binomial distribution Expectation: Variance: Standard deviation:

27 Probabilities Continuous random variable 0°0° Random variable W: angle P: all angles equally likely P(W = 180°) = ? precision: 1°  360 possibilities with P=1/360 precision: 0.1°  3600 possibilities with P=1/3600 precision: 0.01°  possibilities with P=1/36000 n possibilities with P=1/n P(W = 180°) = 0

28 Probabilities Continuous random variable X F 0°0° Distribution function: F(a) = P(W  a) F(0) = P(W  0°) = F(90) = P(W  90°) = F(180) = P(W  180°) = F(360) = P(W  360°) = F(270) = P(W  270°) =

29 Probabilities Continuous random variable X F Distribution function F X f Density function f 1

30 Probabilities Random variable discretecontinuous Expectation Variance

31 Probabilities Random variable X f Density function f Expectation? Conjecture: 180


Download ppt "PROBABILITIES. Probabilities Basic concepts: Random experiment: All possible outcomes have to be known in advance The result of a particular experimen."

Similar presentations


Ads by Google