TS Nguyen Ngoc Rang; Email: rangbvag@yahoo.com; Website: bvag.com.vn; trang:1

PHN PHI NH THC

Trong phn phi chun, bin ngu nhin l bin s lin tc (cn nng, chiu cao,

tr s ng mu..), cn trong phn phi nh thc (Binomial distribution) bin ngu

nhin l bin ri rc (discrete) hoc gi l bin nh phn, kt cc ch c 2 tnh hung

xy ra nh: sng/cht, c bnh/khng bnh.. s phn b nh th no.

V d. Th tung 1 ng xu, ng nhin mi ln tung c kt cc c lp, c

ngha l kt cc ra Sp (S), Nga (N) ln sau khng ty thuc vo kt qu S,N ca

ln tung trc, nh vy xc sut p(N)=1/2 v xc sut p (S)=1/2. Nu ta tung 4 ln

v mun ra mt N, th ta c th d on xc sut c 2 ln N l cao nht, sau

l c 1N hoc 3 N v cui cng l c 4 ln tung u ra N hoc khng c mt N

no c xc sut thp nht.

Nh vy lm sao tnh xc sut c 0,1,2,3,4 mt nga (N) sau 4 ln tung?

Cch tnh 1:

Ta bit xc sut thc ca mt nga l 1/2 (p=0,5) v mt sp cng 1/2 (q=0,5)

Gi k l s ln c mt nga, tnh xc sut vi k=0,1,2,3,4:

SNNS

SNSN

SSSN SSNN NNNS

SSNS NSSN NNSN

SNSS NSNS NSNN

SSSS NSSS NNSS SNNN NNNN

k = 0 1 2 3 4

Biu 11.1 Cc tnh hung xy ra sau 4 ln tung (n=4)

k=0: xc sut khng (k=0) c mt N no l: SSSS= q.q.q.q (qui tc nhn xc

sut) = 0,5x0,5x 0,5x 0,5= 0,5 4 = 0,0625

k=1: c 4 tnh hung c 1 mt N P (k=1) = 4. 0,5 4= 0,25

k=2: c 6 tnh hung c 1 mt N P (k=2) = 6. 0,5 4= 0,375

k=3: 4 tnh hung c 3 mt N P (k=3) =4. 0,54= 0,25

k=4: ch c 1 tnh hung duy nht P (k=4) = 0,54 =0,0625

Cch tnh 2:

Dng nh thc Newton:

TS Nguyen Ngoc Rang; Email: rangbvag@yahoo.com; Website: bvag.com.vn; trang:2

(p+q)4= p4 +4 p3q + 6p2q2 + 4pq3+q4

(0,5+0,5)4= 0,54 +4x0,54 + 6x0,54 + 4x0,54 +0,54

k=0 tng ng s hng th 1: 0,54 k=1 tng ng s hng th 2: 4x0,54 k=2 .. 3: 6x0,54 k=3 .. 4: 4x0,54 k=4 .. 5: 0,54 By gi ta gi X (k=0,1,2) l bin ngu nhin trn trc honh v p l tn s xc

sut c mt nga trn trc tung v v biu phn phi (xem biu 11.2)

Biu 11.2 Phn phi nh thc vi n=4

11.1 Hm phn phi nh thc:

Tng qut ha hm phn phi nh thc c pht biu nh sau:

Mt th nghim vi n ln th c lp

Gi p l xc sut thnh cng (v d. ra mt nga)

Th xc sut ca k ln th nghim thnh cng l:

knknk p)(1pCP p)n,(k,

!!

!

knn

nC

n

k (T hp n chp k)

Nu n=12, dng cng thc tnh:

0,000244(0,5))(12!12!

12!0;12;0;5)P(k 12

0,00292(0,5))(11!12!

12!1;12;0;5)P(k 11

0,01612(0,5))(10!12!

12!2;12;0;5)P(k 10

.

trong k=0,1,3..,n.

TS Nguyen Ngoc Rang; Email: rangbvag@yahoo.com; Website: bvag.com.vn; trang:3

0,2255(0,5))(6!12!

12!6;12;0;5)P(k 6

.

0,000244(0,5))(12!12!

12!12;12;0;5)P(k 12

Cch tnh 3.

Vo phn mm R vi lnh n gin: dbinom (k,n,p)

V d: tnh xc sut vi k=6, n=12 v p=0,5:

>dbinom (6,12,0.5) th P= 0,2255

Biu 11.3 Phn phi nh thc vi n=12

V khi n cng tng dn, phn phi nh thc c dng phn phi chun vi tr s trung

bnh = np v phng sai 2= np(1-p) theo nh lut gii hn trung tm (central limit

theorem).

Biu 11.4 Phn phi nh thc vi n=100

TS Nguyen Ngoc Rang; Email: rangbvag@yahoo.com; Website: bvag.com.vn; trang:4

Nhn vo biu 4 vi n ln (n=100) chng ta thy phn phi nh thc c phn phi

chun vi tr trung binh l =np=100 x 0,5=50 v phng sai 2= 100 x 0,5 x 0,5= 25

hoc lch chun =5.

11.2 ng dng phn phi nh thc:

V d 1. Bit t l suy dinh dng tr em di 5 tui l 20%. Nu chng ta khm

10 tr di 5 tui. Tnh xc sut c 2 em b suy dinh dng.

Nh vy k=2, n=10 v p=0,20

Xc sut c 2 tr b suy dinh dng.

0,300,8(0,2))(8!10!

10!2;10;0;2)P(k

82

hoc tnh trong R:

>dbinom (2,10,0.2)=0,30

Kt qu: Xc sut c 2 em b suy dinh dng l 30%

V d 2. T l suy dinh dng tr em 20% (nh v d trn).Tnh xc sut c t hn

2 em b suy dinh dng.

Nh vy tnh tng xc sut ca: P (k=0)+P(k=1)+ P(k=2)

Hoc dng hm nh thc tch ly (cummulative binomial probability):

tnh P (X 2) :

> pbinom(2,10,0.2)= 0,67

Ti liu tham kho:

1. Nguyn Vn Tun. Hm phn phi nh phn, trong Phn tch s liu v to biu

bng R. Nh xut bn KH v KT, Thnh ph HCM 2007, tr: 47-50.

2. Moore D.S. and McCabe G.P. Discrete random variables, in Introduction to the

Practice of Statistics, 3rd Edition. W.H. Freeman and Company 1999 USA, pp: 313-

317.