Haqiqatan ham, siz haqsiz. Agar biz $\mathbb{P}(A) $ bilan hodisa $A$ sodir bo'lish ehtimoli va $xy$ to'g'ri va faqat $x$ yoki $y$ bo'lsa, biz $$ \mathbb{P}\({xy \; \text{is to'g'ri}\}) = \mathbb{P}(\{x \; \text{is to'g'ri}\} \cup \{y \; \text{is to'g'ri}) $$ ni aniqlagan holda, biz hodisani uchta alohida holatga bo'lishimiz mumkin, ya'ni $x$ to'g'ri, $y to'g'ri yoki $x$ va $y$ ikkalasi ham to'g'ri. Shundan xulosa qilib aytamanki, $\math
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