Theory of Probability and Mathematical Statistics. Probability theory and mathematical statistics The hunter shoots to the first hit

Task 5.

Condition: The device can be assembled from high quality parts and ordinary quality parts. 40% of devices are assembled from high-quality parts.

For a high-quality device, its reliability over time interval t is 0.95; for conventional devices, the reliability is 0.7. The device was tested for time t and worked flawlessly.

Find the probability that it is assembled from high quality parts.

Solution: H 1 - the device is assembled from high quality parts,

H 2 - the device is assembled from parts of ordinary quality.

The probability of these hypotheses prior to experience:

As a result of the experiment, event A was observed - the device worked flawlessly for time t.

The conditional probabilities of this event under the hypotheses H 1 and H 2 are:

We find the probability of the hypothesis H 1 after the experiment:

probability root mean square variance mathematical

Math statistics

Exercise 1.

Condition: Compose the law of distribution of a discrete random variable X, calculate the mathematical expectation, variance and standard deviation of a random variable.

The hunter shoots the game until it hits, but can fire no more than three shots. The probability of hitting each shot is 0.6. Compose the law of distribution of the random variable X - the number of shots fired by the shooter. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Solution: The probability that the number of misses is 0 is 0.6

  • - the probability that the number of misses is equal to 1 is equal to 0.4 0.6 = 0.24 (missed in the first, hit in the second)
  • - the probability that the number of misses is 2 is equal to 0.4 0.4 0.6 = 0.096 (did not hit in the first two, hit in the third)
  • - the probability that the number of misses is 3 is equal to 0.4 0.4 0.4 = 0.064 (did not hit in the first three)

The mathematical expectation is 0 0.6+1 0.24+2 0.096+3 0.064 = 0.624

M(x*x)=0.24 +0.384+0.576=1.2

D(x)=1.2-0.389376=0.810624

Task 2.

Condition: Random value X given by the distribution function F(X).

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One of the most important concepts of probability theory is the concept random variable.

Random called value, which, as a result of tests, takes certain possible values ​​that are not known in advance and depend on random causes that cannot be taken into account in advance.

Random variables are denoted capital letters Latin alphabet X, Y, Z etc. or in capital letters of the Latin alphabet with the right subscript , and the values ​​​​that can take on random variables - in the corresponding small letters of the Latin alphabet x, y, z etc.

The concept of a random variable is closely related to the concept of a random event. Connection with a random event lies in the fact that the acceptance of a certain numerical value by a random variable is a random event characterized by the probability .

In practice, there are two main types of random variables:

1. Discrete random variables;

2. Continuous random variables.

A random variable is a numerical function of random events.

For example, a random variable is the number of points that fell when throwing a dice, or the height of a student randomly selected from the study group.

Discrete random variables are called random variables that take only remote from each other values ​​that can be enumerated in advance.

distribution law(distribution function and distribution series or probability density) completely describe the behavior of a random variable. But in a number of problems it is sufficient to know some numerical characteristics of the quantity under study (for example, its average value and possible deviation from it) in order to answer the question posed. Consider the main numerical characteristics of discrete random variables.

The distribution law of a discrete random variable any ratio is called , establishing a relationship between the possible values ​​of a random variable and their corresponding probabilities .

The law of distribution of a random variable can be represented as tables:

The sum of the probabilities of all possible values ​​of a random variable is equal to one, i.e. .

The distribution law can be represented graphically: on the abscissa axis, the possible values ​​of a random variable are plotted, and on the ordinate axis, the probabilities of these values; the obtained points are connected by segments. The constructed polyline is called distribution polygon.

Example. A hunter with 4 rounds shoots at the game until the first hit or all the rounds are used up. The probability of hitting with the first shot is 0.7, with each subsequent shot it decreases by 0.1. Draw up the law of distribution of the number of cartridges used up by the hunter.


Solution. Since the hunter, having 4 rounds, can make four shots, then the random value X- the number of cartridges used up by the hunter can take the values ​​1, 2, 3, 4. To find the corresponding probabilities, we introduce the events:

- “hit at i- ohm shot”, ;

- “miss at i- th shot”, and the events and are pairwise independent.

According to the condition of the problem, we have:

,

By the multiplication theorem for independent events and the addition theorem for incompatible events, we find:

(hunter hit the target with the first shot);

(hunter hit the target from the second shot);

(hunter hit the target from the third shot);

(the hunter hit the target from the fourth shot or missed all four times).

Verification: - correct.

Thus, the law of distribution of a random variable X looks like:

0,7 0,18 0,06 0,06

Example. A worker operates three machines. The probability that within an hour the first machine will not require adjustment is 0.9, the second is 0.8, the third is 0.7. Draw up a distribution law for the number of machines that will require adjustment within an hour.

Solution. Random value X- the number of machines that will require adjustment within an hour can take the values ​​0.1, 2, 3. To find the corresponding probabilities, we introduce the events:

- “i- th machine will require adjustment within an hour”, ;

- “i- th machine will not require adjustment within an hour”, .

By the condition of the problem, we have:

, .