The formula of probability theory
In principle, the study of this topic does not taketoo much time. In order to get an answer to the question: "How to find the probability of some phenomenon?", You need to understand the key concepts and remember the basic principles on which the calculation is based. So, according to statistics, the events under investigation are denoted by A1, A2, ..., An. Each of them has both favorable outcomes (m) and the total number of elementary outcomes. For example, we are interested in how to find the probability that an even number of points will be on the top of the cube. Then A is a die roll, m is a loss of 2, 4 or 6 points (three favorable variants), and n are all six possible variants.
P (A) = m / n.
It is easy to calculate that in our example the desiredthe probability is 1/3. The closer the result to unity, the more likely that such an event will actually happen, and vice versa. Here is a theory of probability.
With one outcome everything is extremely easy.But how to find the probability, if the events go one after another? Consider this example: one card is displayed from a card deck (36 pcs.), Then it is hidden again in the deck, and after mixing the following is pulled out. How to find the probability that at least in one case the lady was rushed out? There is the following rule: if you are considering a complex event that can be divided into several incompatible simple events, you can first calculate the result for each of them, and then add them together. In our case it will look like this: 1/36+ 1/36 = 1/18. But what about when there are severalindependent events occur simultaneously? Then the results are multiplied! For example, the probability that, when two coins are rolled up simultaneously, two tails will drop out, will be: ½ * ½ = 0.25.
Now let's take an even more complicated example. Suppose we hit a book lottery in which out of thirty tickets ten are winning. It is required to determine:
- The likelihood that both will be winning.
- At least one of them will bring a prize.
- Both will be losing.
So, consider the first case.It can be divided into two events: the first ticket will be happy, and the second will also be happy. We will take into account that the events are dependent, since after each pull-out the total number of variants decreases. We get:
10/30 * 9/29 = 0.1034.
In the second case, you will need to determine the probability of a losing ticket and take into account that it can be either the first account or the second: 10/30 * 20/29 + 20/29 *10/30 = 0.4598.
Finally, the third case, when on the raffled lottery, even one book can not be obtained: 20/30 * 19/29 = 0.4368.