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You are not logged in. #1 20130218 11:34:35
Probabilities: pmf and cdfI am reading and rereading the textbook and I still do not understand what exactly is the pmf and cdf? Since there are two possible combinations for 2, 4, and 6 people, we need to take a union of probabilities for these combinations and final P(X) become this: And from here I need to find pmf All examples in the textbook are just doing direct substitution p(x)=P(X=x) but if I summarize my P(X) I receive 1.4575616. And sum of p(x) for all x is supposed to be equal 1.0... So, did I make a mistake somewhere or what am I missing? #2 20130218 12:20:12
Re: Probabilities: pmf and cdfThose probabilities are not correct. The problem is that if you want exactly one person to be late, you only get the probability that one of the two is late, but not that everybody else isn't. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20130219 10:15:50
Re: Probabilities: pmf and cdfI am not sure I understand. If we say that for the "exactly one group is late" we need to calculate: intersection of "one group is late" and "four groups are not late", and any of the five groups can be late. So the formula for the one group become: But once I recalculate whole table like this  the sum become 1.98. #4 20130219 10:34:50
Re: Probabilities: pmf and cdfTry using the exact values. And, also, it should be a 2 there, not a 5. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #5 20130219 14:22:00
Re: Probabilities: pmf and cdfOk... Lets do it this way: And if I collapse rows with same number of people as a union of the two probabilities I receive: Still no 1.0 in the total. What is wrong now? Last edited by White_Owl (20130219 14:22:54) #6 20130219 19:30:23
Re: Probabilities: pmf and cdfHi Last edited by anonimnystefy (20130219 19:35:18) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #8 20130220 05:43:15
Re: Probabilities: pmf and cdfYou're welcome. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment 