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d) faqat bittasi puch bo„lishi ehtimolligini toping.
Yechish: a) ko„rish mumkinki,
(1-urug„ to„q)
Endi agar birinchi olingan urug„ to„q bo„lsa, u holda paketda 9 ta urug„
qoladi va ularning 5 tasi to„q bo„ladi.
Shuning uchun
(ikkala urug„ ham to„q)
(1-urug„ to„q) (2-urug„ to„q, agar 1-
urug„ to„q bo„lsa)
ekanligini topamiz.
b)
‒ ikkala urug„ ham to„q hodisasi bo„lsin. U holda
hodisa
ikkala urug„dan kamida bittasi puch ekanligini bildiradi.
(ikkita urug„dan kamida bittasi puch)
d) ikkita urug„dan bittasi to„q, ikkinchisi puch bo„ladigan hodisani
ikki xil usulda amalga oshirish mumkin. Chunki birinchi urug„ to„q
bo„lib, ikkinchisi puch yoki birinchi urug„ puch bo„lib, ikkinchisi to„q
bo„lishi mumkin. Agar to„q urug„ tanlangan bo„lsa, u holda paketda
beshta to„q va to„rtta puch urug„ qoladi. Shuning uchun
(2-urug„ puch, agar 1-urug„ to„q bo„lsa)
ekanligi kelib chiqadi. Shuning uchun
(1-urug„ to„q va 2-urug„ puch)
(1-urug„ to„q) (2-urug„ puch, agar
1-urug„ to„q bo„lsa)
ekanligini topamiz. Xuddi shuningdek,
(1-urug„ puch va 2-urug„ to„q
)
bo„ladi. Bu ikki hodisa o„zaro birgalikda bo„lmagan
hodisalar, shuning
uchun
(ikkala urug„dan faqat bittasi puch)
Bu topilgan ehtimollikning to„g„riligini
tekshirish uchun quyidagi
ehtimollikni qaraymiz:
(ikkala urug„ ham puch)
(1-urug„ puch) (2-urug„ puch, agar 1-
urug„ puch bo„lsa)
Endi “ikkala urug„ ham to„q”, “faqat bitta urug„ puch” va “ikkala urug„
ham puch” hodisalari o„zaro birgalikda bo„lmagan hodisalardir va
uchalasi birgalikda barcha mumkin bo„lgan holatlarni beradi.
Shuning
uchun ularning ehtimollari yig„indisi 1 ga teng bo„lishi kerak. Buni esa
osongina tekshirib ko„rish mumkin.
