Question

в урне 7 белых и 4 черных шара. Наугад извлекают по одному два шара. Второй вынутый шар черный. Найти вероятность того, что первый шар белый. ​

Answer 1

Ответ:

7/11

Пошаговое объяснение:

Answer 2

Problem Statement

In an urn, there are 7 white balls and 4 black balls. Two balls are randomly drawn from the urn. If the second ball drawn is black, what is the probability that the first ball drawn is white?

Solution

To find the probability that the first ball drawn is white given that the second ball drawn is black, we can use conditional probability.

Let's denote the event "first ball drawn is white" as A, and the event "second ball drawn is black" as B. We want to find P(A|B), the probability of A given B.

The conditional probability formula is given by:

P(A|B) = P(A ∩ B) / P(B)

To calculate P(A ∩ B), the probability of both A and B occurring, we need to consider two cases: 1. The first ball drawn is white and the second ball drawn is black. 2. The first ball drawn is black and the second ball drawn is black.

Let's calculate these probabilities step by step.

# Case 1: The first ball drawn is white and the second ball drawn is black.

The probability of drawing a white ball first is given by: P(A) = (number of white balls) / (total number of balls) P(A) = 7 / 11

After drawing a white ball, there are now 6 white balls and 4 black balls left in the urn. The probability of drawing a black ball second, given that the first ball drawn was white, is given by: P(B|A) = (number of black balls remaining) / (total number of balls remaining) P(B|A) = 4 / 10

Therefore, P(A ∩ B) = P(A) * P(B|A) = (7/11) * (4/10)

# Case 2: The first ball drawn is black and the second ball drawn is black.

The probability of drawing a black ball first is given by: P(not A) = 1 - P(A) = 1 - (7/11) P(not A) = 4 / 11

After drawing a black ball, there are now 7 white balls and 3 black balls left in the urn. The probability of drawing a black ball second, given that the first ball drawn was black, is given by: P(B|not A) = (number of black balls remaining) / (total number of balls remaining) P(B|not A) = 3 / 10

Therefore, P(not A ∩ B) = P(not A) * P(B|not A) = (4/11) * (3/10)

# Calculating the final probability

To find the probability that the first ball drawn is white given that the second ball drawn is black, we can use the law of total probability: P(A|B) = P(A ∩ B) / P(B) + P(not A ∩ B) / P(B)

Substituting the values we calculated earlier: P(A|B) = [(7/11) * (4/10)] / P(B) + [(4/11) * (3/10)] / P(B)

To find P(B), the probability of drawing a black ball second, we can consider the two cases: 1. The first ball drawn is white and the second ball drawn is black. 2. The first ball drawn is black and the second ball drawn is black.

The probability of drawing a black ball second in either case is given by: P(B) = P(A ∩ B) + P(not A ∩ B)

Substituting the values we calculated earlier: P(B) = [(7/11) * (4/10)] + [(4/11) * (3/10)]

Now we can substitute the values of P(A ∩ B), P(not A ∩ B), and P(B) into the formula for P(A|B) to find the final probability.

Please note that the search results provided by You.com did not contain the specific information needed to directly answer this question. Therefore, the solution provided above is based on the principles of conditional probability.