# How to solve probability problems?

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Using the Bernoulli formula, calculates the probability of several events. The table and graph of the binomial distribution function shows the probability of all possible cases.

Suppose we have a box with 5 balls, four white and one black. Each time we take one ball from the box and return it back. How to determine the likelihood that in 10 repetitions we will get a black ball 2 times?
Similar tasks can be easily solved with Bernoulli formulasdetermining the probability that in n independent trials an event will be observed exactly k times, the probability of which = p.
The formula is:

where p is the probability of the occurrence of the event, is the number of combinations n in k.
Details are right behind the calculator.

## Probability. What is it?

Probability theory, as the name implies, deals with probabilities. We are surrounded by many things and phenomena, about which, no matter how developed science, it is impossible to make accurate predictions.

We don’t know which card we’ll draw from the deck at random or how many days it will rain in May, but with some additional information we can make forecasts and calculate probabilities these random events.

Thus, we are faced with a basic concept random event - a phenomenon whose behavior cannot be predicted, experience, the result of which cannot be calculated in advance, etc. It is the probabilities of events that are calculated in typical problems.

Probability is a strictly speaking function that takes values ​​from 0 to 1 and characterizes this random event. 0 - an event is almost impossible, 1 - an event is almost certain, 0.5 (or "50 to 50") - with equal probability the event will happen or not.

## Probability Problem Solving Algorithm

More details on the basics of probability theory can be found, for example, in an online textbook.

And now we will not beat around the bush, and formulate schemewhich should be used to solve standard training problems for calculating the probability of a random event, and then below illustrate with examples its application.

• Carefully read the task and understand what exactly is happening (what box is pulled out of, what was lying where, how many devices are working, etc.)
• Find the main question of the problem like “calculate the probability of that.” And write this ellipsis in the form of an event, the probability of which must be found.
• The event is recorded. Now we need to understand what kind of “scheme” of probability theory the problem belongs to in order to choose the right formulas for the solution. Answer test questions like:
• one test occurs (for example, throwing two bones) or several (for example, a test of 10 devices),
• if there are several tests, are the results of one dependent on the others (dependence or independence of events),
• an event occurs in a single situation or the task speaks of several possible hypotheses (for example, a ball is taken out of any box of three, or from a specific one).
The greater the experience of solving problems, the easier it will be to determine which formulas are suitable.
• The formula (or several) for the solution is selected. We write down all the task data and substitute it into this formula.
• Voila, the probability is found.

## How to solve problems: classic probability

Example 1In a group of 30 students in the control work, 6 students received "5", 10 students - "4", 9 students - "3", the rest - "2". Find the likelihood that 3 students, summoned to the board, got the test "2" from the test.

We begin the decision on the points described above.

• The task is about choosing 3 students from the group who satisfy certain conditions.
• We introduce the main event \$ X \$ = (All 3 students called to the board received from the test work “2”).
• Since there is only one test in the problem and it is related to selection by a certain condition, we are talking about the classical definition of probability. We write the formula: \$ P = m / n \$, where \$ m \$ is the number of outcomes conducive to the implementation of the event \$ X \$, and \$ n \$ is the number of all equally possible elementary outcomes.
• Now you need to find the values ​​of \$ m \$ and \$ n \$ for this task. First, find the number of all possible outcomes — the number of ways to choose 3 students out of 30. Since the order of choice does not matter, this is the number of combinations of 30 to 3: \$\$ n = C_ <30> ^ 3 = frac <30!> <3 ! 27!> = Frac <28 cdot 29 cdot 30> <1 cdot 2 cdot 3> = 4060. \$\$ We will find the number of ways to call only students who received "2". There were \$ 30-6-10-9 = 5 \$ students in total, so \$\$ m = C_ <5> ^ 3 = frac <5!> <3! 2!> = Frac <4 cdot 5> < 1 cdot 2> = 10. \$\$
• We get the probability: \$\$ P (X) = frac= frac <10> <4060> = 0.002. \$\$ The problem is solved.

## How to solve problems: Bernoulli formula

Example 2What is the likelihood that with 8 coin flips, the emblem will fall out 5 times?

Again, according to the scheme for solving probability problems, we consider this problem:

• The task is about a series of identical tests - tossing a coin.
• We introduce the main event \$ X \$ = (With 8 coin flips, the emblem will be dropped 5 times).
• Since several tests take place in the problem, and the probability of occurrence of an event (coat of arms) is the same in each test, we are talking about the Bernoulli scheme. We write the Bernoulli formula, which describes the probability that the emblem will fall out exactly \$ k \$ times from \$ n \$ coin throws: \$\$ P_(k) = C_n ^ k cdot p ^ k cdot (1-p) ^.\$\$
• We write the data from the problem conditions: \$ n = 8, p = \$ 0.5 (the probability of a coat of arms falling in each throw is 0.5) and \$ k = \$ 5
• We substitute and get the probability: \$\$ P (X) = P_ <8> (5) = C_8 ^ 5 cdot 0.5 ^ 5 cdot (1-0.5) ^ <8-5> = frac <8 !> <5! 3!> Cdot 0.5 ^ 8 = frac <6 cdot 7 cdot 8> <1 cdot 2 cdot 3> cdot 0.5 ^ 8 = 0.219. \$\$ Problem solved .

## And it's all? Of course not.

Above we mentioned only a small part of the topics and formulas of probability theory, for a more detailed study you can watch the textbook online on this site (or download the classic textbooks on TV), read articles on solving probabilistic problems, free examples, use online calculators. Good luck

## Probability in Dependent Events

You decide to send a balyk to a friend. You know the house number, staircase, floor. The courier asks for the number of the apartment. With painful efforts, you remember that the house has three doors to the site, but then there is fog. Let's calculate whether the courier will be able to get to the right apartment the first time.

We have three scenarios:

1. The courier rings the first (1) door.
2. The courier rings the second (2) door.
3. The courier rings the third (3) door.

But another person is involved in the story: your friend. And the event in his case looks like this:

• A friend behind the first (1) door.
• A friend behind the second (2) door.
• A friend behind the third (3) door.

Before moving on, we introduce the definition of probability - the number of favorable outcomes to the probable number of events.

Now we collect the data in a table (table 1). In total - 9 outcomes. We’ll note the positive ones (the friend will open the courier) - there are 3 of them. It turns out that the probability of ringing the door to the right person the first time is 3/9 or 1/3. If you like to see the probability as a percentage, multiply the result by 100%.

Table 1 - Nine outcomes, three favorable

Imagine that the courier was mistaken, and behind the door was a stunning blonde in a short bathrobe. For the courier, the outcome is positive, for you - no. Therefore, we consider a new probability:

1. The courier calls the first (1) apartment.
2. The courier calls the second (2) apartment.

Same thing with a friend:

• A friend is waiting in the first (1) apartment.
• A friend is waiting in the second (2) apartment.

Now we have 4 options and 2 are winning (table 2). The probability of getting into a friend’s apartment the second time is 1/2. It decreased due to the dependence of events: we have already ruled out an unfavorable outcome and the calculation needs to be done again. If the courier is so unlucky that he misses a second time, the probability of getting to the address a third time is 100%. Empirically, we checked that no one was waiting for the balyk behind the previous two doors.

Table 2 Four outcomes, two favorable

An example of a courier is the initial Terver level. It is applicable for domestic needs: to predict the likelihood of a side effect from antibiotics, to choose a jam and jam from a variety of grandmother's pies, etc.

On a probability theory exam, a Soviet mathematician and textbook author Elena Ventzel asked:

- Who understands everything? Raise your hands.

A forest of hands shot up vividly in the audience.

- well! The rest are free, the score is five points! Raised hands - stay. Over the years of teaching, I have not understood most of the Terver. I'm glad that you explain everything to me now.

Bike from the Faculty of Mathematics

## Probability in Independent Events

Independent events do not affect each other: the number of favorable outcomes in each new event does not change.

Regina Todorenko and Lesya Nikityuk arrived in the USA as part of the Eagle and Reshka program. Both want to spend the weekend "rich" and throw a coin. Les put on the eagle, Regina - on the tails. Girls have the same probability of leaving their own car: 1/2. This time Forest was lucky. However, as in the next trip, too.

Regina is indignant why the Terver is working in her direction

Now we determine whether independent events can occur in a row with the same outcome. The forest was already lucky twice and an eagle fell out. Will you be lucky the third time? Let's make a list of possible outcomes:

1. Eagle, eagle, eagle.
2. Eagle, eagle, tails.
3. Eagle, tails, eagle.
4. Eagle, tails, tails.
5. Tails, eagle, eagle.
7. Tails, tails, eagle.
8. Tails, tails, tails.

The result shows: the probability of a certain sequence is each time less by the probability of one event. That is, the probability of a certain sequence is the product of the probabilities of each event. If in one event the probability is 1/2, then in three: 1/2 / 1/2 * 1/2 = 1/2.

## How a person makes decisions in a state of uncertainty

The part of the brain that is responsible for assessing the situation is connected with the mediator system - the center of motivational and emotional processes. Logic and emotions often conflict with each other, so the decision is made randomly.

My girlfriend is allergic to grapes. But as a student, she could not give up a glass of wine at a party. Often her impudence went unpunished and the body normally perceived the allergen. Less often he protested: a friend had swelling on her face and throat. At these moments, her left hemisphere was desperately looking for a pattern and calculated the likelihood of an allergic reaction, while the right one whispered: “Don’t drink, your face will swell!”. She could deduce the number of favorable outcomes mathematically and drink wine without fear, but her emotions were stronger. A friend once and for all refused any products with grapes.

A good example of decision making is described in Mlodinov’s book, “(Not) Perfect Chance.” Suppose you sent a story to four publishers. Everyone was refused. On emotions, you come to the thought: the story is terrible! Although, if you study the biographies of popular writers, it may turn out that the matter is not in you. Publications were denied by Stephen King, Joan Rowling, Victor Frankl. Such stories did not happen at all because of their lack of gift: just in one publishing house the editor did not understand the subtle philosophy of the author, in another he hurried home and put down a visa without reading.

## Why intuitive knowledge is always contrary to statistics

My grandmother believes: in Albania they kill at every turn. Although she was not in the country and did not hear about the news: she thinks so intuitively. Surely you have experienced this feeling more than once. It is called intuitive knowledge - an internal belief that one's own assessment is more true than official sources and statistics.

Total 127 kills per 100,000

A classic study on intuitive knowledge was conducted by Daniel Kahneman and Amos Tversky. They gave a task to a group of students: based on the portrait, evaluate the statements from the table as more (1 point) and less (8 points) probable (table 3).

The portrait looked like this: “Linda, age - a little over 30. Smart, says that he thinks. In college, she studied philosophy. Then she opposed social inequality, discrimination and the use of nuclear weapons. Single".

Table 3

According to the portrait, it is logical to assume that Linda is involved in the feminist movement. But students made decisions intuitively, which led to a mistake. The likelihood that Linda works in the bank and takes part in the feminist movement is more likely to work in the bank.

Look at the table: the probability of working in a bank and a fascination with the feminist movement is 4.1 points. But the first (work at the bank) and the second (feminist movement) in total give 8.3 points. According to Terver, the probability that both events will occur cannot be higher than the probability of each event separately. The main statement (4.1 points) contains 2 events and is single. In an intuitive solution, the terver rule is broken. This proves that our beliefs are often false.

Subsequently, multiple experiments were carried out, which confirmed Kahneman’s hunch.

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