Question

Перед вами три ларца.В первом ларце лежат две золотые монеты,во втором-две серебрянные,а в третьем-одна золотоя и одна серебрянная.Вы вибираете наугад один из ларцов,и наугад вытягиваете оттуда одну монету.Она оказывается золотая.Какая вероятность того,что вторая монета тоже будет золотая?

Answer 1

С золотой монетой есть 2 варианта:
золотая+золотая
золотая+серебряная
Значит вероятность того, что вторая монета будет золотой равна 0,5

Answer 2

Вероятность : 40 %•••••

Answer 3

Problem Analysis

In this problem, we are given three boxes. The first box contains two gold coins, the second box contains two silver coins, and the third box contains one gold coin and one silver coin. We randomly choose one box and randomly draw one coin from it. We are told that the drawn coin is gold, and we need to find the probability that the second coin in the box is also gold.

Solution

To solve this problem, we can use conditional probability. Let's denote the events as follows: - A: The first coin drawn is gold. - B: The second coin in the box is gold.

We need to find the probability of event B given event A, denoted as P(B|A).

Using conditional probability, we have the formula: P(B|A) = P(A and B) / P(A)

To find P(A and B), we need to consider the possible scenarios: 1. We choose the first box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 2/2 = 1, and the probability of the second coin being gold is also 1. 2. We choose the second box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 0/2 = 0, and the probability of the second coin being gold is also 0. 3. We choose the third box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 1/2 = 0.5, and the probability of the second coin being gold is also 0.5.

Therefore, P(A and B) = (1/3) * 1 + (1/3) * 0 + (1/3) * 0.5 = 0.5/3 = 1/6.

To find P(A), we need to consider the possible scenarios: 1. We choose the first box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 2/2 = 1. 2. We choose the second box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 0/2 = 0. 3. We choose the third box: The probability of this event is 1/3. In this case, the probability of drawing a gold coin is 1/2 = 0.5.

Therefore, P(A) = (1/3) * 1 + (1/3) * 0 + (1/3) * 0.5 = 1.5/3 = 1/2.

Now we can calculate P(B|A): P(B|A) = P(A and B) / P(A) = (1/6) / (1/2) = (1/6) * (2/1) = 1/3.

Answer

The probability that the second coin is also gold, given that the first coin drawn is gold, is 1/3.