there are ten students stand before you ,each student has unique ID in range between 1 to 10,let three students whose ID are 1,2,3 seperately Stay together and other seven students stay without order, what probablity does that arrangement happen? Above all, the three students can be regard as a part of those student , all students can be placed in eight position include that part and the other seven students; so the number of permutations with that arrangement is 8!3!8!3!8!3! and the number of whole samples is 10!10!10!. according to the classical probability,the probability of that arrangement is 8!3!10!\frac {8!3!}{10!}10!8!3!. let four studens stay together,so the probablity is 8!3!10!\frac {8!3!}{10!}10!8!3!. let two students stay together and three students stay together,how to compute the probability in that situation. firstly,the two students and three ones are considered as two teams respectively and the other five students make five teams separately. secondly, the two students have 2!2!2! permutations and the three ones have 3!3!3! permutations. 7!3!2!10!\frac {7!3!2!}{10!} 10!7!3!2!
there are 45 ball in the box, including 20 red ball,12 green ball,5 blue ball,8 black ball.11 balls will be took out of them,what probability of geting 6 red balls,2 green balls , 3 blue balls? C206C122C53C4511\frac {C_{20}^{6}C_{12}^{2}C_{5}^{3}} {C_{45}^{11}}C4511C206C122C53
if every sample in a sample space AAA which made by an experiment BBBhappens with same probability and the sample space AAA is an infinited range among straight line or a measure area in plane and space,then the BBB can be called as geometric random experiment, the corresponding model of probability is geometric model.
