Linnyk

Hey again, thanks for helping me! Been working some more on this problem with a friend and we decided E(X)= (40^2)/148 + (33^2)/148 + (25^2) + (50^2)/148 = 39,28 The number seems reasonable. However, I personally don't understand why we square the numbers, could anyone explain this for me? (Is it even correct to do so?) For the busdriver we simply calculated E(Y)= 40/4 + 33/4 + 25/4 + 50/4 = 37

Hi! I'm trying to work on a problem but it seems I'm really stuck and don't even know where to start. 148 students ride in 4 buses, which transport 25, 33, 40 and 50 students respectively. A student is chosen at random. Let X be the number of student on the bus she's riding. Also, one of the 4 busdrivers are chosen at random. Let Y be the number of students on his bus. Calculate expected values E[X] and E[Y]. (I've calculated expected values before but this question seems different and I'm really at a loss as to where to start.) Would really love some help with this!

Ok. I see now that I've obviously missed one important thing, and that is using my results from a) - the probability of passing at least two exams, as that is "given" in B... But I'm not sure what to do. I need to divide it I think. I'm thinking the answer is 1 - (( B * C * Ac ) / 0.674) = 0.938 (Calculating the probability that he's not passed the math exam and subtracting this from 1 (the sum of all possible outcomes) My head is beginning to spin... Am I getting it right?

Hi Timo, thanks for your reply! YES, I realized now there are four ways to pass at least two exams: he may also pass them all This is what I've got now (I've renamned to math to A, physics to B and biology to C): P(passing at least two exams) = A * B * Cc + A * C * Bc + B * C * Ac + A* B * C = 0.8 * 0.7 * 0.7 + 0.8 * 0.3 * 0.3 + 0.7 * 0.3 * 0.2 + 0.8 * 0.7 *0.3 = 0.674 And for b ) (Given that he has passed two of the exams, what is the probability that he has passed the exam in Math?) I'm thinking the anser is 1 - ( B * C * Ac ) = 1 – 0.042 = 0.958 (This seems a high number though.) Does this seem right? (fingers crossed )

Hi! I have a question about probablility. I don't know what the name of this type of statistics is, but here goes: John has three exams coming up, one in Math, one in Physics and one in Biology. He thinks his chances of passing are 0.8 for Math, 0.7 for Physics and 0.3 for Biology. His result on each exam is independent of the others. a) What is the probablility that he passes at least two of the exams? b ) Given that he has passed two of the exams, what is the probability that he has passed the exam in Math. I would love some help with this, either in the form of a direction to online pages that has examples of this kind of problems with solutions, or if you could aid me with what type of formula I could use. (Sorry, might be writing in "Swenglish" here, but I hope I hope you understand what I mean.) The way I see it there are three possible ways he could pass two exams (M+P, M+B and P+B ). Could I for example multiply the probabilility M*P*B-complement (and for the other two respectively, and then add the answers toghether (for a))? Any help would be much appreshiated!

Hi everyone! I have a question; is there an ethological explination for suicide? I was primarily thinking of suicide in humans, but other animals are of interest as well (though I don't think any other animals actually comit suicide?)... In my human ethology course literature ("Human evolutionary psychology") I tried to look up "suicide" in the index, but it seems the book doesn't deal with the subject at all. So, how does one deal with such things as suicide (which certainly isn't uncommon) in the field of human ethology? Comitting suicide at an early age certainly doesn't maximize an individual's fitness... Thanks in advance for any thoughts on the matter! /Linda