King Arthur has two equally wise advisers – Merlin and Percival. Each of them finds the correct answer to any question with probability \(p\) or an incorrect answer, with probability \(q = 1 - p\).
If both counsellors say the same thing, the king listens to them. If they say opposite things, then the king chooses a solution by tossing a coin.
One day, Arthur thought about why he had two advisers, would one not be enough? Then the king called for his counsellors and said:
“It seems to me that the probability of making the right solutions will not decrease if I keep one adviser and listen to him. If so, I must fire one of you. If not, I’ll leave it as it is. Tell me, should I fire one of you?”.
“Who exactly are you going to fire, King Arthur?”, asked the advisers.
“If I make the solution to fire one of you, I will make a choice by tossing a coin”.
The advisers went to think about the answer. The advisors, we repeat, are equally wise, but not equally honest. Percival is very honest and will try to give the right answer, even if he faces dismissal. And Merlin, honest about everything else, in this situation decides to give such an answer with which the probability of his dismissal is as low as possible. What is the probability that Merlin will be fired?
On the skin of a Rhinoceros, its folds are vertical and horizontal. If the Rhinoceros has \(a\) vertical and \(b\) horizontal folds on the left side, and on the right side – \(c\) vertical and \(d\) horizontal folds, we will say that this is a rhinoceros in the state \((abcd)\) or just an \((abcd)\) rhinoceros.
If the Rhinoceros’ itches one of his sides against a tree in an up-down movement, and Rhinoceros has two horizontal folds on this side, then these two horizontal folds are smoothed out. If there are no two folds like this, then nothing happens.
Similarly, if the Rhinoceros itches on of his sides in a back and forth movement, and on this side, there are two vertical folds, then they are smoothed out. If there are no two folds like this, then nothing happens.
If, on some side, two folds are smoothed out, then on the other side, two new folds immediately appear: one vertical and one horizontal.
The rhinoceroses often have random sides that are itchy and need to be scratched against a tree in random directions.
At first there was a herd of Rhinoceroses in the savannah \((0221)\). Prove that after some time there were Rhinoceros of state \((2021)\) in the savannah.
At the sound of the whistle of the PE teacher, all 10 boys and 7 girls lined up randomly.
Find the mathematical expectation of the value “the number of girls standing to the left of all of the boys.”
Hercules meets the three-headed snake Hydra of Lerna. Every minute, Hercules chops off one head of the snake. Let \(x\) be the survivability of the snake (\(x > 0\)). The probability \(p_s\) of the fact that in the place of the severed head will grow s new heads \((s = 0, 1, 2)\) is equal to \(\frac{x^s}{1 + x + x^2}\).
During the first 10 minutes of the battle, Hercules recorded how many heads grew in place of each chopped off one. The following vector was obtained: \(K = (1, 2, 2, 1, 0, 2, 1, 0, 1, 2)\). Find the value of the survivability of the snake, under which the probability of the vector \(K\) is greatest.
In his laboratory, the Scattered Scientist created a unicellular organism, which, with a probability of 0.6 is divided into two of the same organisms, and with a probability of 0.4 dies without leaving any offspring. Find the probability that after a while the Scattered Scientist will not have any such organisms.
Hercules meets the three-headed snake, the Lernaean Hydra and the battle begins. Every minute, Hercules cuts one of the snake’s heads off. With probability \(\frac 14\) in the place of the chopped off head grows two new ones, with a probability of \(1/3\), only one new head will grow and with a probability of \(5/12\), not a single head will appear. The serpent is considered defeated if he does not have a single head left. Find the probability that sooner or later Hercules will beat the snake.
Anna is waiting for the bus. Which event is most likely?
\(A =\{\)Anna waits for the bus for at least a minute\(\}\),
\(B = \{\)Anna waits for the bus for at least two minutes\(\}\),
\(C = \{\)Anna waits for the bus for at least five minutes\(\}\).
In the set \(-5\), \(-4\), \(-3\), \(-2\), \(-1\), \(0\), \(1\), \(2\), \(3\), \(4\), \(5\), replace one number with two other integers so that the set variance and its mean remain unchanged.
Henry wrote a note on a piece of paper, folded it two times, and wrote “FOR MOM” on the top. Then he unfolded the note, added something to it, randomly folded the note along the old folding lines (not necessarily in the same way as he did it before) and left it on the table with random side up. Find the probability that “FOR MOM” is still on the top.
Alice has six magic pies in her pocket: two magnifying pies (if you eat it, you will grow), and two reducing pies (if you eat it, you will shrink). When Alice met Mary Ann, she, without looking, took out three pies from her pocket and gave them to Mary Ann. Find the probability that one of the girls does not have any magnifying pies.