Binomske naključne spremenljivke
V tej objavi bomo razpravljali o binomskih naključnih spremenljivkah.
Predpogoj: Naključne spremenljivke
Posebna vrsta diskretna naključna spremenljivka, ki šteje, kako pogosto se določen dogodek zgodi v določenem številu poskusov ali poskusov.
Da je spremenljivka binomska naključna spremenljivka, morajo biti izpolnjeni VSI naslednji pogoji:
- Obstaja določeno število poskusov (fiksna velikost vzorca).
- Pri vsakem poskusu se dogodek, ki nas zanima, zgodi ali ne.
- Verjetnost pojava (ali ne) je pri vsakem poskusu enaka.
- Preizkušnje so neodvisne ena od druge.
Matematični zapisi
n = number of trials
p = probability of success in each trial
k = number of success in n trials
Zdaj poskušamo ugotoviti verjetnost k uspeha v n poskusih.
Tukaj je verjetnost uspeha v vsakem poskusu p neodvisna od drugih poskusov.
Torej najprej izberemo k poskusov, v katerih bo uspeh, v ostalih n-k poskusih pa neuspeh. Število načinov za to je
Ker je vseh n dogodkov neodvisnih, je verjetnost k uspeha v n poskusih enakovredna množenju verjetnosti za vsak poskus.
Tukaj je njegovih k uspehov in n-k neuspehov. Torej je verjetnost za vsak način doseganja k uspeha in n-k neuspehov
Zato je končna verjetnost
(number of ways to achieve k success
and n-k failures)
*
(probability for each way to achieve k
success and n-k failure)
Potem je verjetnost binomske naključne spremenljivke podana z:
Naj bo X binomska naključna spremenljivka s številom poskusov n in verjetnostjo uspeha v vsakem poskusu p.
Pričakovano število uspehov je podano z
E[X] = np
Varianca števila uspehov je podana z
Var[X] = np(1-p)
Primer 1 : Razmislite o naključnem poskusu, v katerem se 10-krat vrže pristranski kovanec (verjetnost glave = 1/3). Poiščite verjetnost, da bo število prikazanih glav 5.
rešitev:
Let X be binomial random variable
with n = 10 and p = 1/3
P(X=5) = ?
Tukaj je izvedba za isto
C++
Java// C++ program to compute Binomial Probability #include#include using namespace std ; // function to calculate nCr i.e. number of // ways to choose r out of n objects int nCr ( int n int r ) { // Since nCr is same as nC(n-r) // To decrease number of iterations if ( r > n / 2 ) r = n - r ; int answer = 1 ; for ( int i = 1 ; i <= r ; i ++ ) { answer *= ( n - r + i ); answer /= i ; } return answer ; } // function to calculate binomial r.v. probability float binomialProbability ( int n int k float p ) { return nCr ( n k ) * pow ( p k ) * pow ( 1 - p n - k ); } // Driver code int main () { int n = 10 ; int k = 5 ; float p = 1.0 / 3 ; float probability = binomialProbability ( n k p ); cout < < 'Probability of ' < < k ; cout < < ' heads when a coin is tossed ' < < n ; cout < < ' times where probability of each head is ' < < p < < endl ; cout < < ' is = ' < < probability < < endl ; } Python3// Java program to compute Binomial Probability import java.util.* ; class GFG { // function to calculate nCr i.e. number of // ways to choose r out of n objects static int nCr ( int n int r ) { // Since nCr is same as nC(n-r) // To decrease number of iterations if ( r > n / 2 ) r = n - r ; int answer = 1 ; for ( int i = 1 ; i <= r ; i ++ ) { answer *= ( n - r + i ); answer /= i ; } return answer ; } // function to calculate binomial r.v. probability static float binomialProbability ( int n int k float p ) { return nCr ( n k ) * ( float ) Math . pow ( p k ) * ( float ) Math . pow ( 1 - p n - k ); } // Driver code public static void main ( String [] args ) { int n = 10 ; int k = 5 ; float p = ( float ) 1.0 / 3 ; float probability = binomialProbability ( n k p ); System . out . print ( 'Probability of ' + k ); System . out . print ( ' heads when a coin is tossed ' + n ); System . out . println ( ' times where probability of each head is ' + p ); System . out . println ( ' is = ' + probability ); } } /* This code is contributed by Mr. Somesh Awasthi */C## Python3 program to compute Binomial # Probability # function to calculate nCr i.e. # number of ways to choose r out # of n objects def nCr ( n r ): # Since nCr is same as nC(n-r) # To decrease number of iterations if ( r > n / 2 ): r = n - r ; answer = 1 ; for i in range ( 1 r + 1 ): answer *= ( n - r + i ); answer /= i ; return answer ; # function to calculate binomial r.v. # probability def binomialProbability ( n k p ): return ( nCr ( n k ) * pow ( p k ) * pow ( 1 - p n - k )); # Driver code n = 10 ; k = 5 ; p = 1.0 / 3 ; probability = binomialProbability ( n k p ); print ( 'Probability of' k 'heads when a coin is tossed' end = ' ' ); print ( n 'times where probability of each head is' round ( p 6 )); print ( 'is = ' round ( probability 6 )); # This code is contributed by mitsJavaScript// C# program to compute Binomial // Probability. using System ; class GFG { // function to calculate nCr // i.e. number of ways to // choose r out of n objects static int nCr ( int n int r ) { // Since nCr is same as // nC(n-r) To decrease // number of iterations if ( r > n / 2 ) r = n - r ; int answer = 1 ; for ( int i = 1 ; i <= r ; i ++ ) { answer *= ( n - r + i ); answer /= i ; } return answer ; } // function to calculate binomial // r.v. probability static float binomialProbability ( int n int k float p ) { return nCr ( n k ) * ( float ) Math . Pow ( p k ) * ( float ) Math . Pow ( 1 - p n - k ); } // Driver code public static void Main () { int n = 10 ; int k = 5 ; float p = ( float ) 1.0 / 3 ; float probability = binomialProbability ( n k p ); Console . Write ( 'Probability of ' + k ); Console . Write ( ' heads when a coin ' + 'is tossed ' + n ); Console . Write ( ' times where ' + 'probability of each head is ' + p ); Console . Write ( ' is = ' + probability ); } } // This code is contributed by nitin mittal.PHP< script > // Javascript program to compute Binomial Probability // function to calculate nCr i.e. number of // ways to choose r out of n objects function nCr ( n r ) { // Since nCr is same as nC(n-r) // To decrease number of iterations if ( r > n / 2 ) r = n - r ; let answer = 1 ; for ( let i = 1 ; i <= r ; i ++ ) { answer *= ( n - r + i ); answer /= i ; } return answer ; } // function to calculate binomial r.v. probability function binomialProbability ( n k p ) { return nCr ( n k ) * Math . pow ( p k ) * Math . pow ( 1 - p n - k ); } // driver program let n = 10 ; let k = 5 ; let p = 1.0 / 3 ; let probability = binomialProbability ( n k p ); document . write ( 'Probability of ' + k ); document . write ( ' heads when a coin is tossed ' + n ); document . write ( ' times where probability of each head is ' + p ); document . write ( ' is = ' + probability ); // This code is contributed by code_hunt. < /script>Izhod:
Probability of 5 heads when a coin is tossed 10 times where probability of each head is 0.333333
is = 0.136565Ustvari kviz
