[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 516
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I looked it up. You're friends with infinity, but you're just in love with statistics ... That's why it's wrong. There is no limit in the conditions of the problem to Petya's right to choose colours from among the colours of the balls in the bag.
Take your time, drknn. Justify your reasoning.
We get confused because of the inaccurate wording of the problem condition. Or more precisely because of the inaccurately worded question.
1. There are three colours. Petya guessed one of these colours, and Vasya randomly chose a random colour from among them. The probability that Vasya guessed a colour = 1/3, because the question is what is the probability of getting the right colour. In other words, the question is ONE EVENT = a fall-out.
2. There are three colours. Petya guessed one of these colours, and Vasya randomly chose a random colour from among them. If we change the question and ask what is the probability of both events happening at the same time, the answer will change. The model solution will be two three-sided dice, one in Petya's hands and one in Vasya's. And the question will be reformulated as follows: what is the probability that both dice will roll the same number?
The answer is self-evident - we are facing a tuple. The total number of combinations = 3 to the power of 2 = 9 possibilities. The only three winning combinations are 1-1, 2-2, 3-3. Therefore the probability of simultaneous falling out = 3/9 = 1/3.
3 There are three colours. Petya has guessed one of those colours, and Vasya, to choose a random colour, used a random generator - an urn, containing 4 balls - 2 white, 1 blue and 1 red. What is the probability of ONE event occurring - a correctly guessed ball?
2 white + 1 blue and 1 red = 4 balls. The probability of white falling out = 2/4 = 1/2 = 50%. Probability of guessing blue = probability of guessing red = 1/4 = 25%.
4 There are three colours. Petya guessed one of these colours, and Vasya, to choose a random colour, used a random generator - an urn, in which there are 4 balls - 2 white, 1 blue and 1 red. What is the probability of two coinciding events - coincidence of colours in both participants?
Once again we have a motorcade in front of us. It consists of two disks (like the disks with numbers on an electric meter). The first disc, smaller in diameter, has just three digits. The second disc is larger in diameter than the first and has, say, number 1 occurring twice, and numbers 2 and 3 only once. Calculate the number of possible combinations. 3 combinations of the first disk multiplied by 4 combinations of the second disk. That makes 12 combinations. So as not to confuse the counting of combinations we denote the second one as 1' - just to distinguish it from the first one.
Winning combinations: 1-1, 1-1', 2-2, 3-3. Total 4 combinations out of 12 possible. Probability of a match = 4/12 = 1/3.
Vladimir, Petya riddled himself with pink.
No, of course, there is no restriction: Petya chooses the colours just as randomly as Vasya does (by turning away from Vasya and pulling a ball out of the opaque bag).
OK, I'll look again.
Alexei, can you estimate the probability in this case?
That's out of the question, there's no pink.
I've already assessed it and I don't see the error yet.
You can go through all the options. I'll try. But it's too much like brute force.
Read the classics:
There are two white, one blue and one red balloons.
Petya has puzzled over a colour.
Vasya tries to guess the given colour at random. What is the probability that Vasya will guess?