Question

Имеются две урны. В первой 7 белых и 7 черных шаров, во второй 8 белых и 6 черных шаров. Из второй урны в первую перекладывают один шар. После этого из первой урны извлекают один шар. Он белый. Какова вероятность того, что шар, переложенный из второй урны в первую, был чёрным?

Answer 1

Решим по формуле полной вероятности

пусть событие А-из первой урны вынут черный шар. результат выбора зависит от цвета шаров, которые перекладывали из второй урны.

Может возникнуть две  гипотезы: Н₁-из 2 -й урны переложили белый шар, Н₂-из 2-й урны переложили  черный шар,

Эти события несовместны, хоть одна из них. да произойдет, они образуют полную группу событий.

Р(Н₁)= 8/14=4/7; всего шаров во второй урне 14.

аналогично найдем Р(Н₂)= 6/14=3/7;

Контроль Р(Н₁)+Р(Н₂)=(4/7)+(3/7)=1

Найдем теперь условные вероятности, что вытащили черный из первой урны,  при условии, что туда переложили соответственно белый или черный шар из второй урны.

P(A|H₁)=7/15; добавили белый. в первую урну, их стало 8, а всего 15

P(A|H₂)=8/15 - добавили черный, их стало 8, а всего 15.

используем формулу полной вероятности.

P(A)=P(H₁)P(A|H₁)+P(H₂)P(A|H₂)=(4/7)*(7/15)+(3/7)*(8/15)=(28+24)/105=52/105

Ответ 52/105

Answer 2

Problem Analysis

We have two urns: the first urn contains 7 white and 7 black balls, and the second urn contains 8 white and 6 black balls. One ball is transferred from the second urn to the first urn, and then one ball is drawn from the first urn. We want to find the probability that the transferred ball was black, given that the drawn ball from the first urn is white.

Solution

To solve this problem, we can use conditional probability. Let's denote the events as follows: - A: The transferred ball is black. - B: The drawn ball from the first urn is white.

We want to find P(A|B), the probability that the transferred ball was black given that the drawn ball is white.

We can use Bayes' theorem to calculate this probability: ``` P(A|B) = (P(B|A) * P(A)) / P(B) ``` where: - P(B|A) is the probability of drawing a white ball from the first urn given that the transferred ball was black. - P(A) is the probability that the transferred ball was black. - P(B) is the probability of drawing a white ball from the first urn.

Let's calculate each of these probabilities step by step.

1. P(A): The probability that the transferred ball was black. - In the second urn, there are 6 black balls out of a total of 14 balls (8 white + 6 black). - Therefore, the probability of transferring a black ball is 6/14.

2. P(B|A): The probability of drawing a white ball from the first urn given that the transferred ball was black. - After transferring one ball from the second urn to the first urn, the total number of balls in the first urn is 15 (7 white + 8 black). - Therefore, the probability of drawing a white ball from the first urn, given that the transferred ball was black, is 7/15.

3. P(B): The probability of drawing a white ball from the first urn. - The total number of balls in the first urn is 15 (7 white + 8 black). - Therefore, the probability of drawing a white ball from the first urn is 7/15.

Now, let's substitute these values into Bayes' theorem to calculate P(A|B): ``` P(A|B) = (P(B|A) * P(A)) / P(B) = (7/15 * 6/14) / (7/15) = 6/14 = 3/7 ```

Answer

The probability that the ball transferred from the second urn to the first urn was black, given that the drawn ball from the first urn is white, is 3/7.