Ehtimolligining

partiyadan ikkitasida yutishnimi;	Ikki partiyadan bir partiyada yutishi		To‘rt partiyadan ikkitasida yutishi		Ikkkalasi teng			To’g’ri javob yo’q
5.
Muavr-Laplasning lokal formulasini ko’rsating.	P k  ^1 ^n n pq	x	   k e Pn k k!		Pn k1; k2  x" x			^'^P k  ^n^!p^k q^n ^n k!n  k !		k 6.
Agar biror hodisaning ro‘y berish ehtimoli 0,4 ga teng bo‘lsa, bu hodisaning 100 ta sinovdan rosa 50 marta ro’y berish ehtimolini Muavr- Laplas formulasi yordamida toping.	0,0102		0,5000		0,2010			0,0201		7.
Muavr-Laplasning integral formulasini ko’rsating.	Pn k1; k2  x" x		'Pn k  	n! k!n  k !	p^k^Pqn ^n^^k^^k^^	1 n pq	x	Pn k  	k e k!		8.
Tanga 5 marta tashlanadi. «Gerbli» tomon kamida ikki marta tushish ehtimolini toping.	13 16		1 12		3 16			8 9		9.
Laplas funksiyasinig toq yoki juft ekanligini ayting.	Toq		Juft		Toq ham emas, juft ham emas			To’g’ri javob yo’q		10.
Darslik 100000 tirajda bosib chiqarilgan. Darslikning varaqlari noto’g’ri yig’ilgan bo’lish ehtimoli 0,0001 ga teng. Butun tirajda rosa beshta brak kitob bo’lish ehtimolini toping.	0,0375		0,336		0,815			0,0102		11.

Bernulli formulasidan foydalanib hodisaning k dan kam marta ro’y berish ehtimolini ko’rsating.	Pn 0 Pn 1 ...  Pn k1 	1Pn 0 Pn 1 ...  Pn 	kPn (k)  Pn k 1 ...  Pn n	Pn k 1 Pn k  2 ...  Pn 	n 12.
Zavod bazaga 500 ta buyum jo’natdi. Yo’lda buyumning shikastlanish eytimoli 0,002 ga teng. Yo’lda rosa 3 ta buyumning shikastlanish ehtimolini toping.	0,0613	0,336	0,0375	0,0102	13.
Bernulli formulasidan foydalanib hodisaning k dan ko’p marta ro’y berish ehtimolini ko’rsating.	Pn k 1 Pn k  2 ... 	Pn Pnn (k)  Pn k 1 ...  Pn n	Pn 0 Pn 1 ...  Pn k	Pn 0 Pn 1 ...  Pn k1 1	14.
Agar bitta sinovda A hodisaning ro’y berish ehtimoli 0,4 ga teng bo’lsa, u holda to’rtta erkli sinovda A hodisaning kamida uch marta ro’y berish ehtimolini toping.	0,1792	0,0613	0,0375	0,0102	15.
Bernulli formulasidan foydalanib hodisaning kamida k marta ro’y berish ehtimolini ko’rsating.	Pn (k)  Pn k 1 ...  Pn n	Pn 0 Pn 1 ...  Pn 	k1 Pn 1k1 Pn k  2 ...  Pn	nPn 0 Pn 1 ...  Pn k	16.
Oilada 5 farzand bor. Bu bolalar orasida ikkitadan ortiq o’g’il bolalar bo’lish ehtimolini toping. O’g’il bolalar tug’ilish ehtimolini 0,51 ga teng deb oling.	0,52	0,38	0,76	0,68	17.
Bernulli formulasidan foydalanib hodisaning ko’pi bilan k marta ro’y berish ehtimolini ko’rsating.	Pn 0 Pn 1 ...  Pn k	Pn (k)  Pn k 1 ...  Pn n	Pn 0 Pn 1 ...  Pn k1 	1Pn k 1 Pn k  2 ...  Pn n	18.
Har bir sinovda ro`y berish ehtimoli ^P^^A^^^p^(0^^p^^1)ga teng bo`lgan A hodisaning n ta bog`liq bo`lmagan tajribalar ketma-ketligida ro`y berishlar soni k ning k1 , k2  kesmada (yopiq oraliqda) bo`lishi ehtimoli taqriban	Laplas	Eyler	Bernulli	Puasson	19.

Pn (k1  k  k2 )  (x2 )  (x1) ga teng. Bu yerda – qanday funksiya?
Tasodifiy hodisaning har bir sinovda ro`y berish ehtimoli ^P^^A^^^p^(0^^p^^1)ga teng bo`lsin. ta bog`liq bo`lmagan tajribalar ketma-ketligida etarlicha katta bo’lsa, hodisaning k marta ro`y berish ehtimoli hisoblash formulasini ko’rsating.	P k   ^1x ^n n pq	  k e Pn k  k!	P k   ^n^!p^k q^n ^n k!n  k !	Tk o’g’ri javob yo’q	20.

Tasodifiy miqdorlar. Diskret tasodifiy miqdorlar. Taqsimot qonuni

Test topshirig’i						To’g’ri javob						Muqobil javob				Muqobil javob					Muqobil javob
Quyidagi jadvallardan qaysi biri taqsimot qonuni bo’la oladi?						xi pi		 2 0,1	1 0,4	1 0,3	2 0,2	xi pi		0 1 0,1 0,3	2 0,1	3^xi ^0^,^3^pi		1 0 1 1 4 2	1 2 1 1 4 4	xi pi			1 3 0,4 0,3	5 0,1	1.
Tasodifiy miqdorlar uchun qaysilari to’g’ri?						Diskret va uzluksiz						Muqarrar va muqarrar bo’lmagan				Tasodifiy, ishonchli, ishonchsiz					Tasodifiy, diskret, uzlluksiz				2.
Quyidagi jadval taqsimot qonunni ifoda qilish uchun х qanday qiymatga ega bo’lishi lozim?						0,2						0,4				0,5					0,3				3.
	xi	1	3	5
	pi	x	0,3	0,5

X diskret tasodifiy miqdor-tangani ikki marta tashlashda “gerb”li tomon tushish sonining binomial taqsimot qonunini yozing.					xi Pi		0 1 4	1 1 2	2 1 4	xi Pi			0 1 4	1 1 4	2 1 4	3 1 4	xi Pi			0 3 7	1 2 7	2 2 7	xi Pi		0 1 2	1 1 2		2 1 2					4.
X tasodifiy miqdor- 4 ta talabadan iborat guruhdagi iqtidorli talabalar soni. Bu miqdorning mumkin bo’lgan qiymatlarini yozing.					^x1=0, ^x2=1, ^x3=2, ^x4=3, ^x5=4						^x1=1, ^x2=2, ^x3=3, ^x4=4, ^x5=5							^x1=1, ^x2=2, ^x3=3, ^x4=4						^x1=0, ^x2=1, ^x3=2, ^x4=3										5.