1. Muammo: Ikki zavoddan keltirilgan maxsuIotlar orasida tanlangan maxsuIot sifatsiz bo'lsa, u ikkinchi zavoddan bo'lish ehtimolini topish. 2. Formulalar va tushunchalar: Bayes teoremasi yordamida hisoblaymiz: $$P(B|A) = \frac{P(A|B)P(B)}{P(A)}$$ Bu yerda: - $A$ — maxsuIot sifatsiz bo'lish hodisasi - $B$ — maxsuIot ikkinchi zavoddan bo'lish hodisasi 3. Ma'lumotlar: - $P(B) = 0.45$ - $P(B^c) = 0.55$ (birinchi zavod) - $P(A|B) = 0.3$ - $P(A|B^c) = 0.1$ 4. $P(A)$ ni topamiz: $$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c) = 0.3 \times 0.45 + 0.1 \times 0.55 = 0.135 + 0.055 = 0.19$$ 5. Bayes teoremasi bo'yicha: $$P(B|A) = \frac{0.3 \times 0.45}{0.19} = \frac{0.135}{0.19} \approx 0.7105$$ 6. Javob: Sifatsiz maxsuIot ikkinchi zavoddan bo'lish ehtimoli taxminan 0.7105 yoki 71.05%. 1. Muammo: 120 marta tosh tashlandi. Toshda tushgan raqamlar yig'indisi juft bo'lish ehtimoli $p=0.5$. Juft yig'indisi kamida 65 va ko'pi bilan 75 ta bo'lish ehtimolini toping. 2. Bu binomial taqsimot muammosi: $X \sim Binomial(n=120, p=0.5)$, $P(65 \leq X \leq 75)$ ni topamiz. 3. Binomial ehtimollikni hisoblash uchun: $$P(65 \leq X \leq 75) = \sum_{k=65}^{75} \binom{120}{k} (0.5)^k (0.5)^{120-k} = \sum_{k=65}^{75} \binom{120}{k} (0.5)^{120}$$ 4. Hisoblash murakkabligi sababli, normal taqsimotga yaqinlashtirish mumkin: $$\mu = np = 120 \times 0.5 = 60$$ $$\sigma = \sqrt{np(1-p)} = \sqrt{120 \times 0.5 \times 0.5} = \sqrt{30} \approx 5.477$$ 5. Normal taqsimotga o'tkazamiz va diskretlikni tuzatamiz: $$P(65 \leq X \leq 75) \approx P(64.5 < Y < 75.5)$$ $$Z_1 = \frac{64.5 - 60}{5.477} \approx 0.82$$ $$Z_2 = \frac{75.5 - 60}{5.477} \approx 2.83$$ 6. Standart normal taqsimot jadvalidan: $$P(Z < 2.83) \approx 0.9977$$ $$P(Z < 0.82) \approx 0.7939$$ 7. Natija: $$P(65 \leq X \leq 75) \approx 0.9977 - 0.7939 = 0.2038$$ 8. Javob: Juft yig'indisi 65 dan 75 gacha bo'lish ehtimoli taxminan 0.204 yoki 20.4%.