Binomial distribution
Probability mass function
Cumulative distribution function
الترميز
B
(
n
,
p
)
{\displaystyle B(n,p)}
الوسائط
n
∈
{
0
,
1
,
2
,
…
}
{\displaystyle n\in \{0,1,2,\ldots \}}
p
∈
[
0
,
1
]
{\displaystyle p\in [0,1]}
الحامل
k
∈
{
0
,
1
,
…
,
n
}
{\displaystyle k\in \{0,1,\ldots ,n\}}
PMF
(
n
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle {\binom {n}{k}}p^{k}(1-p)^{n-k}}
CDF
I
1
−
p
(
n
−
k
,
1
+
k
)
{\displaystyle I_{1-p}(n-k,1+k)}
المتوسط
n
p
{\displaystyle np}
أوسط
⌊
n
p
⌋
{\displaystyle \lfloor np\rfloor }
⌈
n
p
⌉
{\displaystyle \lceil np\rceil }
منوال
⌊
(
n
+
1
)
p
⌋
{\displaystyle \lfloor (n+1)p\rfloor }
⌈
(
n
+
1
)
p
⌉
−
1
{\displaystyle \lceil (n+1)p\rceil -1}
تباين
n
p
(
1
−
p
)
{\displaystyle np(1-p)}
تخالف
1
−
2
p
n
p
(
1
−
p
)
{\displaystyle {\frac {1-2p}{\sqrt {np(1-p)}}}}
تدبب زائد
1
−
6
p
(
1
−
p
)
n
p
(
1
−
p
)
{\displaystyle {\frac {1-6p(1-p)}{np(1-p)}}}
الاعتلاج
1
2
log
2
(
2
π
e
n
p
(
1
−
p
)
)
+
O
(
1
n
)
{\displaystyle {\frac {1}{2}}\log _{2}\left(2\pi enp(1-p)\right)+O\left({\frac {1}{n}}\right)}
shannons . For nats , use the natural log in the log. MGF
(
1
−
p
+
p
e
t
)
n
{\displaystyle (1-p+pe^{t})^{n}}
CF
(
1
−
p
+
p
e
i
t
)
n
{\displaystyle (1-p+pe^{it})^{n}}
PGF
G
(
z
)
=
[
(
1
−
p
)
+
p
z
]
n
{\displaystyle G(z)=[(1-p)+pz]^{n}}
معلومات فيشر
g
n
(
p
)
=
n
p
(
1
−
p
)
{\displaystyle g_{n}(p)={\frac {n}{p(1-p)}}}
n
{\displaystyle n}
Binomial distribution for
p
=
0.5
{\displaystyle p=0.5}
with
n and
k as in
Pascal's triangle The probability that a ball in a
Galton box with 8 layers (
n = 8) ends up in the central bin (
k = 4) is
70
/
256
{\displaystyle 70/256}
.
توزيع احتمالي ثنائي هو توزيع لتجربة عشوائية لها ناتجان فقط أحدهما نجاح التجربة والآخر فشلها ويكون الشرط الأساسي أن احتمال النجاح لا يتأثر بتكرار التجربة ، أمثلة : رمي قطعة نقود ، الإحصاءات أو الأسئلة التي تعتمد الإجابة لا أو نعم.
بتعبير آخر التوزيع الاحتمالي ثنائي الحد هو تكرار لتجربة برنولي (انظر توزيع برنولي ).
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
خصائص التوزيع الثنائي يتميز التوزيع الثنائى بعدة خصائص هي:
تتكون التجربة من أكثر من محاولة. إذا تكونت التجربة من محاولة واحدة ،فإننا في تجربة توزيع برنولي
استقلال المحاولات عن بعضها البعض أي ثبات احتمال النجاح p ومن ثم احتمال الفشل q.
هذه المحاولات جميعا متماثلة ومستقلة.
احتمال النجاح ثابت في كل محاولة. قالب:بعض التوزيعات الاحتمالية الشائعة بمتغير واحد
F
(
k
;
n
,
p
)
=
Pr
(
X
≤
k
)
=
I
1
−
p
(
n
−
k
,
k
+
1
)
=
(
n
−
k
)
(
n
k
)
∫
0
1
−
p
t
n
−
k
−
1
(
1
−
t
)
k
d
t
.
{\displaystyle {\begin{aligned}F(k;n,p)&=\Pr(X\leq k)\\&=I_{1-p}(n-k,k+1)\\&=(n-k){n \choose k}\int _{0}^{1-p}t^{n-k-1}(1-t)^{k}\,dt.\end{aligned}}}
Some closed-form bounds for the cumulative distribution function are given below .
Example Suppose a biased coin comes up heads with probability 0.3 when tossed. What is the probability of achieving 0, 1,..., 6 heads after six tosses?
Pr
(
0
heads
)
=
f
(
0
)
=
Pr
(
X
=
0
)
=
(
6
0
)
0.3
0
(
1
−
0.3
)
6
−
0
=
0.117649
{\displaystyle \Pr(0{\text{ heads}})=f(0)=\Pr(X=0)={6 \choose 0}0.3^{0}(1-0.3)^{6-0}=0.117649}
Pr
(
1
heads
)
=
f
(
1
)
=
Pr
(
X
=
1
)
=
(
6
1
)
0.3
1
(
1
−
0.3
)
6
−
1
=
0.302526
{\displaystyle \Pr(1{\text{ heads}})=f(1)=\Pr(X=1)={6 \choose 1}0.3^{1}(1-0.3)^{6-1}=0.302526}
Pr
(
2
heads
)
=
f
(
2
)
=
Pr
(
X
=
2
)
=
(
6
2
)
0.3
2
(
1
−
0.3
)
6
−
2
=
0.324135
{\displaystyle \Pr(2{\text{ heads}})=f(2)=\Pr(X=2)={6 \choose 2}0.3^{2}(1-0.3)^{6-2}=0.324135}
Pr
(
3
heads
)
=
f
(
3
)
=
Pr
(
X
=
3
)
=
(
6
3
)
0.3
3
(
1
−
0.3
)
6
−
3
=
0.18522
{\displaystyle \Pr(3{\text{ heads}})=f(3)=\Pr(X=3)={6 \choose 3}0.3^{3}(1-0.3)^{6-3}=0.18522}
Pr
(
4
heads
)
=
f
(
4
)
=
Pr
(
X
=
4
)
=
(
6
4
)
0.3
4
(
1
−
0.3
)
6
−
4
=
0.059535
{\displaystyle \Pr(4{\text{ heads}})=f(4)=\Pr(X=4)={6 \choose 4}0.3^{4}(1-0.3)^{6-4}=0.059535}
Pr
(
5
heads
)
=
f
(
5
)
=
Pr
(
X
=
5
)
=
(
6
5
)
0.3
5
(
1
−
0.3
)
6
−
5
=
0.010206
{\displaystyle \Pr(5{\text{ heads}})=f(5)=\Pr(X=5)={6 \choose 5}0.3^{5}(1-0.3)^{6-5}=0.010206}
Pr
(
6
heads
)
=
f
(
6
)
=
Pr
(
X
=
6
)
=
(
6
6
)
0.3
6
(
1
−
0.3
)
6
−
6
=
0.000729
{\displaystyle \Pr(6{\text{ heads}})=f(6)=\Pr(X=6)={6 \choose 6}0.3^{6}(1-0.3)^{6-6}=0.000729}
[1] Mean If X ~ B (n , p ), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:[2]
E
[
X
]
=
n
p
.
{\displaystyle \operatorname {E} [X]=np.}
For example, if n = 100, and p = 1/4, then the average number of successful results will be 25.
Proof: We calculate the mean, μ , directly calculated from its definition
μ
=
∑
i
=
0
n
x
i
p
i
,
{\displaystyle \mu =\sum _{i=0}^{n}x_{i}p_{i},}
and the binomial theorem :
μ
=
∑
k
=
0
n
k
(
n
k
)
p
k
(
1
−
p
)
n
−
k
=
n
p
∑
k
=
0
n
k
(
n
−
1
)
!
(
n
−
k
)
!
k
!
p
k
−
1
(
1
−
p
)
(
n
−
1
)
−
(
k
−
1
)
=
n
p
∑
k
=
1
n
(
n
−
1
)
!
(
(
n
−
1
)
−
(
k
−
1
)
)
!
(
k
−
1
)
!
p
k
−
1
(
1
−
p
)
(
n
−
1
)
−
(
k
−
1
)
=
n
p
∑
k
=
1
n
(
n
−
1
k
−
1
)
p
k
−
1
(
1
−
p
)
(
n
−
1
)
−
(
k
−
1
)
=
n
p
∑
ℓ
=
0
n
−
1
(
n
−
1
ℓ
)
p
ℓ
(
1
−
p
)
(
n
−
1
)
−
ℓ
with
ℓ
:=
k
−
1
=
n
p
∑
ℓ
=
0
m
(
m
ℓ
)
p
ℓ
(
1
−
p
)
m
−
ℓ
with
m
:=
n
−
1
=
n
p
(
p
+
(
1
−
p
)
)
m
=
n
p
{\displaystyle {\begin{aligned}\mu &=\sum _{k=0}^{n}k{\binom {n}{k}}p^{k}(1-p)^{n-k}\\&=np\sum _{k=0}^{n}k{\frac {(n-1)!}{(n-k)!k!}}p^{k-1}(1-p)^{(n-1)-(k-1)}\\&=np\sum _{k=1}^{n}{\frac {(n-1)!}{((n-1)-(k-1))!(k-1)!}}p^{k-1}(1-p)^{(n-1)-(k-1)}\\&=np\sum _{k=1}^{n}{\binom {n-1}{k-1}}p^{k-1}(1-p)^{(n-1)-(k-1)}\\&=np\sum _{\ell =0}^{n-1}{\binom {n-1}{\ell }}p^{\ell }(1-p)^{(n-1)-\ell }&&{\text{with }}\ell :=k-1\\&=np\sum _{\ell =0}^{m}{\binom {m}{\ell }}p^{\ell }(1-p)^{m-\ell }&&{\text{with }}m:=n-1\\&=np(p+(1-p))^{m}\\&=np\end{aligned}}}
History This distribution was derived by James Bernoulli . He considered the case where p = r /(r + s ) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p = 1/2.
See also الهامش
^ Hamilton Institute. "The Binomial Distribution" October 20, 2010.
^ See Proof Wiki
^ Mandelbrot, B. B., Fisher, A. J., & Calvet, L. E. (1997). A multifractal model of asset returns. 3.2 The Binomial Measure is the Simplest Example of a Multifractal
مراجع
Discrete univariate Discrete univariate Continuous univariate Continuous univariate Continuous univariate Continuous univariate Mixed continuous-discrete univariate Multivariate (joint) Directional Degenerate and singular Families
![{\displaystyle p\in [0,1]}](https://www.marefa.org/api/rest_v1/media/math/render/svg/33c3a52aa7b2d00227e85c641cca67e85583c43c)
![{\displaystyle G(z)=[(1-p)+pz]^{n}}](https://www.marefa.org/api/rest_v1/media/math/render/svg/ad93fbb1208a6637ccd78540a9ddfc2e65a91ee2)
![{\displaystyle \operatorname {E} [X]=np.}](https://www.marefa.org/api/rest_v1/media/math/render/svg/3f16b365410a1b23b5592c53d3ae6354f1a79aff)
