P( X)=M / M+N ; P(not z)= 1-P(Z) P( T/X)=B/M; P(X/T) = B/ B+D “Care e Pr. Sa extragem x daca am extras numai din categoria t?” P( X and T )=(M/N+M) *( B/M) P(X and Y)= (M/M+N) * (N/ M+N) P( X or T )=(M/M+N) +( B+D/M+N) - P(X and T) P(X or Y) =(M/M+N) + (N/ M+N) - P( X and Y)
P(A) = # times A observed/ Total # outcomes (N). A tree diagram -no. of outcomes of events and the sample space. 1) 2) 3) 4)
X Y
Z A C A+C
T B D A+D
A+B=M C+D=N M+N
Folosim N=n*m =>goes back into the pool. Avem m ways of doing A and n ways of doing B. How many possibilities? Folosim N=n! if the outcome after is chose does not go back into the pool.In how many ways we can arrange x things? X! Daca trebuie sa alegem din n k si conteaza ordinea. How many possibilities? Pnk =n! / (n-k)! Daca trebuie sa alegem din n k si NU conteaza ordinea. How many possibilities? Cnk = n! / ((n-k)!* k!)
Relative Frequency = Frequency of A/ Total Frequency = P(A). While statistics (proportions-population) - past events, Probability (sample) - future events. 1) A U B, A or B P(A or B) = P(A) + P(B) - P(A and B) Mutually exclusive => P(A and B)=0 2) P(B/A) Independent=P(B). Dependent=P(B/A)= P(A and B)/ P(A) 3) Intersection A ∩ B ///Dependent :P(A and B) = P(A) x P(B|A) Independent Events: P(A and B) = P(A) x P(B) 4) A stands for “not A” ; Complement Rule: P(A) + P(not A) = 1; P(A) = 1- P(not A); P(At least one) = 1 – P(none)
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1)UNIFORM
DISTRIBUTION: The area under the uniform distribution: P( Mx1 ) z1= 1 - CI% x% (confidence interval dat) of the observation fall below X 3) BINOMIAL DISTRIBUTION: calculez : p(success); n= total no; x = number of successes in sample- ni se da in intrebare; BINOMIAL FORMULA: p(x)=
