Question

Problem 1. A point is randomly thrown inside a circle with radius R. What is the probability that the point belongs to

a) inscribed square;
b) inscribed regular triangle?

Problem 5. We have a box with 10 white, 5 red and 7 green balls. We draw a ball at random, fix the color and then return to the box. We repeat this experiment 10 times.
i) What is the probability that we will have exactly 6 white balls in 10 experiments?

ii) What is the probability that the number of the white balls in 10 experiments will be between 5 and 7 (including 5 and 7)?

iii) What is the probability that we will have exactly 3 white, 6 red and 1 green balls 1 drawn?

iv) What is the probability that we will have at least 6 white, and at least 3 red balls drawn?


Problem 6. At a party 8 men take off their hats. The hats are then mixed up, and each man randomly selects one. We say that a match occurs if a man selects his own hat. What is the probability of no matches?


Problem 7. We select 13 cards from a deck of 52 playing cards. Compute probability that this set of 13 cards is void in at least one suit.

Problem 8. A line segment of length L is being broken in two randomly chosen points. What is the probability that the obtained three pieces can be used to construct a triangle?

Problem 9. Toyota sells millions of vehicles every year worldwide. 92% of Toyota cars are of perfect quality, 7% has a small defect with airbags, and only 1% has a dangerous defect with brakes. If a company purchases a Toyota car 5 times a year, then what is the probability that
a) only 3 vehicles were of perfect quality?
b) 2 vehicles were of perfect quality, 2 had a problem with airbags, and 1 had defective brakes?

Problem 10. What is the probability that the roots of equation x 2 + 2ax + b = 0
a) are real;
b) are positive; if the coefficients a and b are randomly selected from the domain |a| ≤ 1, |b| ≤ 1.

Answer 1

Problem 1: a) To determine the probability that a randomly thrown point belongs to the inscribed square, we need to compare the areas of the square and the circle. The side length of the inscribed square is equal to the diameter of the circle, which is 2R.

The area of the square is (2R)^2 = 4R^2. The area of the circle is πR^2.

Therefore, the probability that the point belongs to the inscribed square is: P(a) = Area of the square / Area of the circle = 4R^2 / (πR^2) = 4/π ≈ 1.273.

b) Similarly, for the inscribed regular triangle, we need to compare the areas of the triangle and the circle. The side length of the equilateral triangle inscribed in the circle is also 2R.

The area of the triangle is (sqrt(3)/4)(2R)^2 = (sqrt(3)/4)(4R^2) = sqrt(3)R^2. The area of the circle is πR^2.

Therefore, the probability that the point belongs to the inscribed regular triangle is: P(b) = Area of the triangle / Area of the circle = (sqrt(3)R^2) / (πR^2) = sqrt(3)/π ≈ 0.551.

Problem 5: i) To calculate the probability of having exactly 6 white balls in 10 experiments, we need to use the concept of binomial probability. Each experiment has a 10/22 chance of drawing a white ball.

The probability of drawing exactly k white balls in n experiments can be calculated using the binomial probability formula:

P(X = k) = (n C k) * (p^k) * ((1 - p)^(n - k))

In this case, n = 10 (number of experiments), k = 6 (number of white balls), and p = 10/22 (probability of drawing a white ball).

Using the formula, we can calculate the probability:

P(6 white balls) = (10 C 6) * ((10/22)^6) * ((1 - 10/22)^(10 - 6))

ii) To calculate the probability that the number of white balls in 10 experiments will be between 5 and 7 (inclusive), we need to find the sum of the probabilities of drawing 5, 6, and 7 white balls.

P(5 or 6 or 7 white balls) = P(5 white balls) + P(6 white balls) + P(7 white balls)

iii) To calculate the probability of having exactly 3 white, 6 red, and 1 green ball in 1 draw, we consider the probability of each color individually and multiply them together.

iv) To calculate the probability of having at least 6 white and at least 3 red balls drawn, we need to consider the probabilities of different combinations of white and red balls, including 6 white and 3 red, 7 white and 3 red, 8 white and 3 red, and so on, up to 10 white and 5 red.

Problem 6: To find the probability of no matches when the hats are randomly selected, we need to consider the number of possible arrangements of the hats.

There are 8 men and 8 hats, so the total number of possible arrangements is 8!.

To count the number of arrangements with no matches, we can use the principle of derangements. A derangement is a permutation of a set in which no element appears in its original