Gli appunti di scienza dei dati[5]

probability

1. there are suppositions that $A$ is random exercise, $G$ is sample space of $A$,every event $E$ of $A$ corresponds with a real number which can be called as probability of $E$,that is $P(E)$.
   the P is regard as a function to tranform an event to the real number belonged to the event. $P(E)$ is ever greater than or equals zero for each event E; $P(E^{\prime })=1$ when the $E^{\prime }$ event necessarily happen; let $E_1,E_2,....$ are events which only one presents at the same time,that is to say, $E_i\cap E_j=\varnothing$.
   $P(E_1\cup E_2\cup ..)=P(E_1)+P(E_2)+...$
   by the way,the following things are characteristics of probability.

• $P(\varnothing )=0$.
• $P(E_1\cup E_2\cup ..\cup E_n)=P(E_1)+P(E_2)+...+P(E_n)$,that is to say it is finite additivity.
• $P(E_2-E_1)=P(E_2)-P(E_1)$
• for a given event ,$P(E)\le 1$
• for arbitrary event $E$, $P(\bar{E})=1-P(E)$,the $\bar{E}$ is inversed event.
  there is an important conclusion that to given both event $E_1$ and $E_2$,$P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1{E}_2)$.
  of course ,the above formula can expand to multiple event’s additivity.for example:
  $P(E_1\cup E_2\cup ..\cup E_n)={\Sigma }_{i=1}^{n}P(E_i)-{\Sigma }_{1\le i<j\le n}P(E_i{E}_j)+{\Sigma }_{i\le i<j<k\le n}P(E_i{E}_j{E}_k)+...+(-1{)}^{n-1}P(E_1{E}_2...E_n)$

2. the classical probability derives from the initial stage of probability theory,that means each one of sample space which has finite samples such as G={G1,G2,...Gn}G=\{G_1,G_2,...G_n\}G={G1​,G2​,...Gn​} has the same possibility of existence.
   to compute a fundamental classical probablity is so simple because that Any two samples in sample space appear mutually exclusively and the finite additivity of probability.for example:
   P(∪i=1n{Gi})=P(G)=1P(\cup_{i=1}^n\{G_i\})=P(G)=1P(∪i=1n​{Gi​})=P(G)=1
   P(∪i=1n){Gi})=Σi=1nP(Gi)=nP(Gi)P(\cup_{i=1}^n)\{G_i\})=\Sigma_{i=1}^nP(G_i)=nP(G_i)P(∪i=1n​){Gi​})=Σi=1n​P(Gi​)=nP(Gi​),$P(G_1)=P(G_2)=...$
   so that,each $P(G_i)=\frac{1}{n}$
   the above example suppose that one event includes only one sample,but in real life,one event perhaps involves some samples,for example,$G_2$ consists of $n_2$ samples,then $P(G_2)=\frac{n_2}{n}$.
3. Permutations and Combinations

• the repetitve permutaion denotes that there are a sets $A$ consisted of $a_{all}$ distinct elements, we take out $a_{part}$ elements from the $A$ sets one by one,achieving total progress costs $a_{part}$ times, that is to say,each time to take one out and put it back.so that, there are ${a_{all}}^{a_{part}}$ kinds of permutations.
• permutation means fetching $a_{part}$ elements from a sets $A$ which includes $a_{all}$ elements to arrange without returning it back ,so that $P_{a_{all}}^{a_{part}}=\frac{a_{all}!}{(a_{all}-a_{part})!}$ kinds of permutations exists in the progress.
• full permutation: when $a_{all}=a_{part}$,$P_{a_{all}}^{a_{part}}=a_{part}!$
• combination: if we get $a_{part}$ elements from $a_{all}$ distinct elements without their order for making a combination ,and do it repeatedly, then the total number of those combinations is as follows.
  $$C_{a_{all}}^{a_{part}}=\frac{P_{a_{all}}^{a_{part}}}{P_{a_{part}}^{a_{part}}}=\frac{a_{all}!}{a_{part}!(a_{all}-a_{part})!}$$

4. we take out an integer from lots of integers with a length of 12 digits, what is the probability that the first five digits are all different ?
   in the first place,one of the first five digits is an integer ranged from 0 to 9,so we have $10^5$ ways to make the integer which has five digits, those ways construct a sample space.in the second place,$P_{10}^5$ represent the number of permutations that five digits total are distinct because that five different numbers will be choosed in a range between 0 and 9 arbitrarily.
   so that we get the probability which is $\frac{P_{10}^5}{10^5}$.

references

1. 《数学》