1-misol. Qutida n ta bir xil detal bo’lib, shulardan m tasi standart. Tavakkaliga olingan k ta detal orasida kamida bitta standart detal bo’lish ehtimolligini toping.
Yechilishi. “Olingan detallarning ichida kamida bittasi standart” va “olingan detallarning ichida bitta ham standart detal yo’q” hodisalari qarama-qarshi hodisalardir. Birinchi hodisani A orqali, ikkinchisini esa  orqali belgilaymiz.
U holda: P(A)=1-P().
P(A) ni topamiz. n ta detaldan k ta detal olish usullarining jami soni  ga teng. Nostandart detallar soni n-m ga teng; shu detallardan k ta nostandart detalni  ta usul bilan olish mumkin. Shuning uchun olingan k ta detal ichida bitta ham standart detal yo’qligining ehtimolligi
ga teng.
Izlanayotgan ehtimollik quyidagiga teng:
 .
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