Question

Sa se determine probabilitatea ca,alegand un element al multimii A={0.1.2,...,10} ,acesta sa verifice inegalitatea n!<100. Multumesc celor care imi dau raspunsuri !

Answer 1

$4!\ \textless \ 100\ \textless \ 5! \Rightarrow 5 ~cazuri~favorabile:~\{0,1,2,3,4\}. \\ \\ P= \frac{nr.caz.fav.}{nr.caz.pos.} = \frac{5}{11}~(= 0,(45) \%).$

Answer 2

n!<100⇒n∈A;
4!=1·2·3·4=6·4=24;
5!=1·2·3·4·5=6·4·5=24·5=120⇒120>100⇒nu este solutie a problemei;
⇒4!<100⇒nr. cazuri favorabile=[0,1,2,3,4]⇒5;
nr. cazuri posibile=10+1=11;
⇒P=5/11=0,[45][la suta];