donys

Hi guys. ===================================================================== Can you help me with these problems in combinations.I dont know if my solutions are correct. ===================================================================== Problem 1 ========= There are 3 classes,each one with n students.We must select 3 students from the total of 3n students. i)How many are the possible choices? C=combinations The possible choices are: C(n,3)*C(n,0)*C(n,0)*3 + C(n,2)*C(n,1)*C(n,0)*3! + C(n,1)*C(n,1)*C(n,1) ii)How many are the possible choices,if the 3 students belong to the same class? C(n,3)*C(n,0)*C(n,0)*3 iii)How many are the possible choices,if 2 of the 3 students belong to the same class and the third to a different class? C(n,2)*C(n,1)*C(n,0)*3! iv)How many are the possible choices,if the 3 students belong to a different class? C(n,1)*C(n,1)*C(n,1) v)Use your answers i)-iv) to expand C(3n,n) I dont know this one ===================================================================== Problem 2 ========= A number of n students attend the lecture of probabilities.The array with the results of the final exam includes only the names of the students that passed the lesson in descending order.We assume that there arent any students with the same grade.How many are the possible arrays? C(n,1)+C(n,2)*2!+C(n,3)*3!+...+C(n,n)n! ===================================================================== Problem 3 ========= We put the hats of n persons in a box.Then each person randomly chooses a hat from the box.Find the probability that n-2 persons will accurately choose their own hat. I guessed that n-2 persons choose acurately their own hat is the same with: 1 - 2 persons dont take their own hat =[1 - C(n,1)*(n-1)!+C(n-1,1)*(n-2)!] / n! =[1-n!+(n-1)!]n!?????i'm not sure?????? =====================================================================

Hi again...This is my last question for today. I know that i send a lot of threads but...i dont have anyone else to help me. Most of the times i dont send any solutions.I try but sometimes i dont understand and so...i dont have something to write.I post frequently because if i dont give a 50% of correct answers i will not have the chance to write in the final exams...its a stupid rule. My last 2 questions for today are. 1.We must prove that if P(A/B)=1 then P(B compl. / A compl.)=1 i wrote a lot of things but ..i didnt prove nothing : ~) 2.A and B are two events with non zero probabilities. I must show (prove) if the following are i)correct,ii)faulse iii)correct under some conventions. a)if A and B are foreign then they are independent (i dont know if this is the correct word) butif they are foreign P(A[intersect]B)=[empty-set] b)if A and B are independent then they are foreign c)P(A)=P(B)=0.6 A,B are foreign c)P(A)=P(B)=0.6 A,B are independent

Hi guys i know that i post a problem without my own answer but this problem is very difficult for me and i'm very comfused. The exercise says that we must implement the Bayes' Law to solve the exercise.My problem is that i cant distinguish and write down the events ): Problem:In my kitchen i put my 10 forks in the left drawer and my 10 knifes in the right drawer. before my room mate change something: k=knife,f=forks left drawer : (10f,0k) right drawer: (0f,10k)) When my room-mate came he took 2 forks from the left drawer and he put them in the right drawer. after my room mate's alteration: left drawer : (8f,0k) right drawer: (2f,10k) After this action he chooses one piece(knife or fork) from the right drawer and he puts the piece in the left drawer. left drawer : (9f,0k) or (8f,1k) right drawer: (1f,10k) or (2f,9k) After all these changes i come and choose one piece from 1 of the 2 drawers. IF Is given that i hold a knife,what is the probability that i opened the left drawer?

Hi...Can you plz check if my proof is correct? Exercise: A1,A2,.....An are independently events. Prove that : P(A1[union]A2[union]...[union]An) = 1-Πi[element-of]I(1-P(Ai)) note for this (Πi[element-of]I(1-P(Ai)) I={1,2,....n) P([intersect]Ai)= Π P(Ai) for 3 events A1,A2,A3 means: P(A1[intersect]A2)=P(A1)*P(A2) P(A2[intersect]A3)=P(A2)*P(A3) P(A2[intersect]A3)=P(A2)*P(A3) P(A1[intersect]A2[intersect]A3)=P(A1)* P(A2) * P(A3) Now my proof: We know that P([intersect]Ai)= Π P(Ai) if A1,A2,...,An are independent then and the complements are independent P([intersect]Ai)complement = Π P(Aicomplement) P([union](Ai compl) ) = Π(1-P(Ai)) 1-P([union]Ai)= Π(1-P(Ai)) -P([union]Ai)=-1+Π(1-P(Ai)) Finally ... we got our proof P([union]Ai)=1-Πi[element-of]I(1-P(Ai)) Is it correct? And one more.... but i dont know how to prove this: A,B,C are independent We must prove that A and B[union]C are independent too ...?

Can you help me with the following problems plz. I have a course in telecommunications and i have to understand complex numbers first. I can't solve the following exercises: 1) Write in the form z=x+jy the complex number e^e^j ^=exp 2)how i can solve this equation |z+2|=|z-1| and what is the algebraical explanation (z=|z|e^jè|)

Hi guys i have a problem but i dont know how to start solving this can you help? We have three events: A={The midday temperature in Hirakleio is 25C} B={The midday temperature in Athens is 25C} C={The maximum of the two midday temperatures in Hirakleio and Athens is 25[degrees]C} P(A)=0.3 P(B)=0.4 P©=0.2 Find the probability that the minimum temperature between the two midday temperatures is 25C

Hi.. I have an exercise and i want to know if i solved it correct.Can you help? Three people A,B,C throw a coin (First is A,second comes B and finally C) The first that will bring head wins and the game stops. The exercise wants: a)The sample space b)Determinate the following events in the sample space i)A={A WINS} ii)B={B WINS) iii)(A union B)complement H:HEAD T:TAIL a)sample space{H,TH,TTH,TTT} bi)P(A)={H}=1/2 bii)P(B)={TH}=P(Acomplement intersection B)=1/4 biii)(A union B)complement=Acomplement intersection Bcomplement= {TTH,TTH}=1/4 is it corrent?