Mavzular

⭐ — imtihonda eng ko'p ishlatiladigan formulalar.

Klassik ta'rif

  • ⭐ Hodisa AA ning ehtimoli (barcha natijalar teng imkoniyatli bo'lsa):

P(A)=mnP(A) = \frac{m}{n}

bunda nn — barcha mumkin bo'lgan natijalar soni, mm — AA uchun qulay natijalar soni.

  • ⭐ 0≤P(A)≤10 \le P(A) \le 1. Mumkin bo'lmagan hodisa uchun P=0P = 0, muqarrar hodisa uchun P=1P = 1. Javob 11 dan katta yoki manfiy chiqsa — xato.
  • ⚠️ Natijalar teng imkoniyatli bo'lsin. Ikkita tanga tashlanganda "gerb–raqam" va "raqam–gerb" alohida natijalar: jami 44 ta (GG,GR,RG,RRGG, GR, RG, RR), 33 ta emas.

Qarama-qarshi hodisa

  • ⭐ P(Aˉ)=1−P(A)P(\bar{A}) = 1 - P(A).
  • ⭐ "Kamida bittasi" hodisasi uchun: P=1−P(birortasi ham)P = 1 - P(\text{birortasi ham}). Masalan, tanga 44 marta tashlanganda kamida bir marta raqam tushishi: 1−116=15161 - \dfrac{1}{16} = \dfrac{15}{16}.

Qo'shish va ko'paytirish qoidalari

  • ⭐ Birgalikda bo'lmagan ("yoki") hodisalar: P(A yoki B)=P(A)+P(B)P(A \text{ yoki } B) = P(A) + P(B).
  • Birgalikda bo'lishi mumkin bo'lganda: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).
  • ⭐ Mustaqil ("va") hodisalar: P(A va B)=P(A)⋅P(B)P(A \text{ va } B) = P(A) \cdot P(B).
    • Misol: nishonga tekkazish ehtimoli har otishda 35\dfrac{3}{5}. Ikkala otishda ham tegish: 35⋅35=925\dfrac{3}{5} \cdot \dfrac{3}{5} = \dfrac{9}{25}. Kamida bittasida tegish: 1−25⋅25=21251 - \dfrac{2}{5} \cdot \dfrac{2}{5} = \dfrac{21}{25}.
    • Aynan bittasida tegish: 2⋅35⋅25=12252 \cdot \dfrac{3}{5} \cdot \dfrac{2}{5} = \dfrac{12}{25}.
  • Bog'liq hodisalar (masalan, shar qaytarilmaydi): P(A va B)=P(A)⋅P(B∣A)P(A \text{ va } B) = P(A) \cdot P(B \mid A), bunda P(B∣A)P(B \mid A) — AA ro'y bergan bo'lsa, BB ning ehtimoli. Qopda 33 qizil va 22 ko'k shar: ketma-ket ikkalasi qizil: 35⋅24=310\dfrac{3}{5} \cdot \dfrac{2}{4} = \dfrac{3}{10}.

Kombinatorika yordamida

  • Guruh tanlash ("bir vaqtda kk ta olindi"): P=CmkCnkP = \dfrac{C_m^{k}}{C_n^{k}} ko'rinishida. Masalan, 1010 ta sharda 44 tasi qizil, 22 ta olinsa ikkalasi qizil: C42C102=645=215\dfrac{C_4^2}{C_{10}^2} = \dfrac{6}{45} = \dfrac{2}{15}.
  • Bu usul qaytarmasdan ketma-ket olishga teng: yuqoridagi misol 410⋅39=215\dfrac{4}{10} \cdot \dfrac{3}{9} = \dfrac{2}{15}.

Tez-tez uchraydigan holatlar

  • ⭐ Tanga nn marta: barcha natijalar 2n2^n. Aynan kk marta gerb: Cnk2n\dfrac{C_n^k}{2^n}. Masalan, 55 marta, aynan 22 gerb: 1032=516\dfrac{10}{32} = \dfrac{5}{16}.
  • ⭐ Ikki kubik (barcha natijalar 6⋅6=366 \cdot 6 = 36). Yig'indi bo'yicha qulay natijalar:
Yig'indi23456789101112
Natijalar12345654321

Masalan, yig'indi 77 bo'lish ehtimoli 636=16\dfrac{6}{36} = \dfrac{1}{6}. Bir kubik: juft ochko — 36=12\dfrac{3}{6} = \dfrac{1}{2}.

  • Bernulli sxemasi: har urinishda muvaffaqiyat ehtimoli pp (q=1−pq = 1 - p) bo'lsa, nn ta urinishda aynan kk ta muvaffaqiyat:

Pn(k)=Cnk pkq n−kP_n(k) = C_n^k \, p^k q^{\,n - k}

Masalan, p=13p = \dfrac{1}{3}, n=3n = 3, k=2k = 2: 3⋅19⋅23=293 \cdot \dfrac{1}{9} \cdot \dfrac{2}{3} = \dfrac{2}{9}.

Geometrik ehtimollik

  • Natijalar kesma, yuza yoki hajmdagi nuqta bo'lsa: P=qulay qismning o’lchamibutun o’lchamP = \dfrac{\text{qulay qismning o'lchami}}{\text{butun o'lcham}}. Masalan, tomoni 22 bo'lgan kvadratga tasodifiy nuqta tashlansa, unga ichki chizilgan birlik doiraga tushish ehtimoli π⋅1222=π4\dfrac{\pi \cdot 1^2}{2^2} = \dfrac{\pi}{4}.

Tipik xatolar

  • ⚠️ Ehtimollarni qo'shish kerak joyda ko'paytirish (va aksincha): "va" — ko'paytiring, "yoki" — qo'shing.
  • ⚠️ Kubiklarni "bir xil" deb hisoblash: (1,2)(1, 2) va (2,1)(2, 1) ikki xil natija.
  • ⚠️ "Kamida bitta"ni to'g'ridan-to'g'ri sanash; qarama-qarshi hodisadan foydalanish ancha oson.
  • ⚠️ Qaytarib olish va qaytarmasdan olishni aralashtirish: ikkinchi holda shar soni kamayadi.

Mashq qiling

30 ta savol, vaqt bilan, izohlari bilan.

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