Версия для печати темы

Автор: Aiven 12.5.2007, 13:50

В 3 урнах находятся белые и черные шары: в первой - 2 белых и 4 черных, во второй - 3 белых и 5 черных, в третьей - 4 белых и 6 черных. Из первой урны взяли наудачу два шара и переложили во вторую. После этого взяли 2 шара из второй урны и переложили в третью. Наконец, из третьей урны в первую переложили 2 шара. Определить, чему равна вероятность того, что: 1) состав шаров во всех урнах не изменился? 2) состав шаров во всех урнах изменился?

Вычислила вероятность Р(Н1) -из первой урны во вторую перекладывают 2 белых шара =1/15
Р(Н2)- 1белый и 1 чёрный шар =8/15
Р(Н3)- 2 чёрных шара =6/15..
Подскажите, что делать дальше?? Аналогично вычислять вероятности из второй урны в третью?

Автор: venja 12.5.2007, 15:04

Цитата(Aiven @ 12.5.2007, 19:50)

Автор: Aiven 12.5.2007, 18:45

Спасибо. Значит, ответ на 2 вопрос будет= 1-Р(А)?

Автор: Ботаник 13.5.2007, 7:03

Думаю надо делать так, как Aiven изложила в первом сообщении. Всего надо рассмотреть вероятности 3^3 случаев. Вычисления не сложные, просто их много.

Автор: venja 13.5.2007, 7:32

Цитата(Aiven @ 13.5.2007, 0:45)

Автор: Aiven 13.5.2007, 10:08

первую я уже высчитала..
на радостях думала, что со вторым вопросом будет проще...
Принцип наверно будет такой: Р(В)=[1-Р(Н1)]*[1-Р(А1)]*[1-Р(А2)]+......????

Автор: venja 13.5.2007, 10:48

Не знаю.
Быть может все-таки имелся в виду вопрос с "хотя бы"?
Иначе, по-моему, получается слишком сложно.

Автор: Aiven 13.5.2007, 11:16

Я тогда чушь написала… Если нарисовать случаи, то скорее всего будет так:

Р(В)-вероятность того, что состав во всех урнах изменился.
Р(В)=Р(Н1) * [P(B1)*P(B2) + P(B3)*P(B4)] +Р(Н2)*….
Р(Н1)-из первой во вторую переложили 2белых шара.

Р(В1)из второй в третью переложили 1б. и 1ч.шары.
Р(В2)-из третьей в первую переложили 2ч.шара

Р(В3)-из второй в третью переложили 2 чёрных шара.
Р(В4)-из третьей в первую переложили 1б. и 1ч.шары.

аналогично для Р(Н2) и Р(Н3)
Правильно???

Автор: venja 13.5.2007, 11:31

А ведь кажется верно! Молодец.
Только надо при вычислении этих вероятностей учитывать меняющийся расклад шаров. Поэтому, думаю, корректней записать так:
Р(В)=Р(Н1) * [P(B1/Н1)*P(B2/(Н1*В1)) + P(B3/Н1)*P(B4/(Н1*В3))] +Р(Н2)*….

Автор: Ботаник 15.5.2007, 6:37

Я добил-таки эту задачу, расписав полную группу событий. Для этого нужно было вычислить вероятности 3^1 + 3^2 + 3^3 событий. Для интересующихся даю ссылку на полученные результаты.

http://www.bottanikk.narod.ru/TeorVer/TV.xls (88 кб)

Например у меня получилось Р(состав шаров во всех урнах не изменился)=0,23340
Р(состав шаров во всех урнах изменился)=0,09412

Автор: venja 15.5.2007, 8:51

Если Aiven посчитала по предыдущей схеме, то было бы интересно сравнить ответы.

Автор: Ботаник 15.5.2007, 9:07

ну так надо у неё спросить

Автор: venja 15.5.2007, 9:12

Спрашиваем!

Автор: Aiven 15.5.2007, 14:26

я считала по схеме:Р(А)=Р(Н1)*Р(А/Н1)+Р(Н2)*Р(А/Н2)+Р(Н3)*Р(А/Н3)
Ответ на первый вопрос получился =0,2337
Ответ на второй вопрос:=0,094
Всё совпадает с вашими ответами)))
Большое спасибо за помощь

Автор: zara 31.3.2013, 20:01

Цитата(venja @ 12.5.2007, 15:04)

Автор: venja 1.4.2013, 14:21

Цитата(Aiven @ 12.5.2007, 19:50)

Автор: chocolet1 26.10.2022, 17:28

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Автор: chocolet1 11.11.2022, 14:09

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