Question

В першій коробці знаходиться 5 білих та 8 чорних кульок, в другій – 2 білі та 7 чорних, в третій – 5 білих та 3 чорних. Знавмання вибраної коробки вийнято кульку. Вона виявилась білою. Знайти ймовірність того, що кульку вийнято з другої коробки.

Answer 1

Ответ:

1) 5+8=13 всього в 1 коробки

2) 2+7=9 всього в 2 коробки

3)5+3=8 всього в 3 коробки

Объяснение:

Answer 2

Problem Analysis

We are given three boxes with different numbers of white and black balls. We are told that a ball is randomly chosen from one of the boxes, and it turns out to be white. We need to find the probability that the ball was chosen from the second box.

Solution

To solve this problem, we can use Bayes' theorem. Bayes' theorem allows us to update our prior probability based on new evidence.

Let's define the following events: - A: The ball is chosen from the first box. - B: The ball is chosen from the second box. - C: The ball is chosen from the third box. - W: The ball is white.

We need to find the probability of event B given that event W has occurred, i.e., P(B|W).

According to Bayes' theorem, we have:

P(B|W) = (P(W|B) * P(B)) / P(W)

We can calculate each term in the equation step by step.

Calculation

1. P(W|B): The probability of drawing a white ball from the second box. - In the second box, there are 2 white balls and 7 black balls. - Therefore, P(W|B) = 2 / (2 + 7) = 2/9.

2. P(B): The probability of choosing the second box. - There are three boxes in total, so the probability of choosing the second box is 1/3.

3. P(W): The probability of drawing a white ball. - We can calculate this by considering all possible cases: - Case 1: Choosing the first box and drawing a white ball. - In the first box, there are 5 white balls and 8 black balls. - Therefore, the probability of choosing the first box and drawing a white ball is (5/13) * (1/3) = 5/39. - Case 2: Choosing the second box and drawing a white ball. - We already calculated P(W|B) = 2/9. - The probability of choosing the second box is 1/3. - Therefore, the probability of choosing the second box and drawing a white ball is (2/9) * (1/3) = 2/27. - Case 3: Choosing the third box and drawing a white ball. - In the third box, there are 5 white balls and 3 black balls. - Therefore, the probability of choosing the third box and drawing a white ball is (5/8) * (1/3) = 5/24. - Since these cases are mutually exclusive, we can add their probabilities to get P(W) = (5/39) + (2/27) + (5/24) = 37/117.

Now, we can substitute the calculated values into Bayes' theorem:

P(B|W) = (P(W|B) * P(B)) / P(W) = (2/9) * (1/3) / (37/117) = 26/37 ≈ 0.703

Therefore, the probability that the ball was chosen from the second box, given that it is white, is approximately 0.703.

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