Practical lesson (part 2). Algebra of events. Properties of probabilities
Practical class "Algebra of events. Properties of probabilities»
Let's revise the basic operations concerning events. That is, if A and B are some events for the space of elementary outcomes Ω (omega), then the event A plus B is the sum of events that occurs if and only if one of these events or both occur together.
Event A multiplied by B is the product of events. It occurs if and only if both events have occurred.
The negation of event A is an event that occurs when event A did not occur.
The probability of negation of event A can be calculated as one minus the probability of event A. At the same time, since one of the axioms of probability tells us that the probability of a reliable event is equal to one, the probability of an impossible event is zero.
And the probability of the sum of two events can be calculated as the probability of the first event, plus the probability of the second event, and minus the probability of their product.
These properties will be enough for us to calculate the probability.
Let's review the relationship between events. For each pair of events, you need to determine whether they are incompatible.
The first pair, event A1 is the failure of the TV working in the living room, A2 is the failure of the TV working in the kitchen.
Since both TVs can fail at the same time, these events will be conjoint, i.e. they will not be incompatible.
Event B1 is to hit the target with one shot, event B2 is to miss a shot.
Since you can't both hit and miss with the same shot, events B1 and B2 will be incompatible.
Moreover, event B2 will be the negation of event B1.
Example Three. Two shooters shoot at the target. Event C1 is when at least one of the shooters hit the target, event C2 is when both shooters hit the target. These events will not be incompatible, since if both shooters hit the target, then at least one shooter also hit. In this case, event C2 entails event C1.
The last example: event D1 is the situation when an even number of points fell out, event D2 is when more than two points fell out. If we have an even number of points it means that two, four, or six points are drawn. If you get more than two points, you get three, four, five, or six points. If we have four or six points, both events will occur.
This means that these events will not be incompatible.
That is, they are compatible. Incompatible events are convenient when calculating probability, since the probability of incompatible events product is the probability of an impossible event, which means it is zero.
In this case, the probability of the sum of incompatible events can be calculated as the sum of their probability.
That is, if we know that the events are incompatible, then we don't need to know the probability of the product of these events.
The next relationship between events that we have considered is an independent event. By definition, the events are independent if and only if the probability of the product of these events is equal to the product of their probabilities.
Let us consider the following experience. We throw three coins consecutively. You need to determine whether or not the following events are dependent: A is when we have “heads” on the first coin and event B is when we have at least one “tails”. The space of elementary outcomes in this case is the set of all possible sequences of "heads" and "tails". That is, it will be: "heads" "heads" "heads" "heads" "heads", "tails"; "heads", "tails", "heads"; "heads", "tails", tails; "tails", "heads", "heads"; "tails", "heads", "tails"; "tails", "tails", "heads"; "tails", "tails", "tails".
That is, there will be eight equally likely elementary outcomes.
The event A is favored by those outcomes where the first coin has "heads". There are four of them, which means that the probability of the event A is four-eighths or one-half. The event B, when we need just one tails, is favored by the following elementary outcomes.
In fact, these are all elementary outcomes, except for the case when we have three "heads".
There are seven favorable outcomes, so the probability of the event B is seven-eighths. What is an AB event? This means that the “heads” is on the first coin, and there is at least one "tails".
That is, among the first four cases, the ones marked with a tick will suit us. These three cases are here (they are highlighted in red).
This means that the number of elementary outcomes favorable to the AB event is three.
Thus, the probability of the event A multiplied by the B event is three-eighths.
Let's check whether the events are independent. If the events are independent, then the equation must be satisfied when the probability of the product of these events is equal to the product of their probabilities. Three-eighths equals seven-eighths multiplied by one-half – this equation is incorrect, so events are dependent.
In other words, you can't use this property here.
Therefore, every time you want to use this formula, you need to make sure that the events in question are independent.
Consider the problem. Let each of the two shooters independently fire one shot at the target.
Event A is when the first shooter hit the target, and event B is when the second shooter hit the target.
Let's understand what the events A plus B; AB; and A but not B mean.
A plus B means that the first shooter hit the target or the second shooter hit the target. That is, at least one of the shooters hit the target. Event A multiplied by B means that the first shooter hit the target and the second shooter hit the target. That is, both shooters hit the target. And event A but not B means that the first shooter hit the target, but the second shooter did not hit. Let us find the probability provided that the probability of event A is 0.7 and the probability of event B is 0.8.
Since we assume that the shooters shoot independently of each other, we can write down that the probability of event A multiplied by B is equal to the product of their probability.
That is, we multiply 0.7 by 0.8. We have 0.56. In other words, we have found the probability of the AB event. Based on the property, the probability of event A plus B can be calculated as the probability of event A plus the probability of event B and minus the probability of their product. That is 0,7 + 0,8 - (0,7*0,8) = 0,94.
The event A but not B. Before calculating its probability, it will be convenient to draw the situation using Euler circles. Here is the set of all elementary outcomes (in a square), event A (in a circle), and event B (in a circle).
A but not B is the set of those elementary outcomes, in our case it is one elementary outcome (marked with red lines). Note that event A but not B and event AB (marked with blue lines) are incompatible, which means that the probability of their sum is equal to the sum of their probabilities.
We know that the probability of the product is 0.56, and the sum of these events is nothing more than event A. The probability of event A is 0.7, which means that the probability of event A but not B can be expressed with this equation: event A but not B is equal to 0.14.
In general, in any unclear situations when you need to express an event in some way, it makes sense to draw them using Euler diagrams. They allow you to visualize what is happening.
The last example before solving problems. Let the shooter make two independent shots at the target. That is, he does not get upset at the fact that he missed and does not rejoice at the fact that he hit.
This does not affect the probability that he will hit again. And we can consider each shot independent of the previous event.
At the same time (let’s draw a target), we know that the probability of making the bull’s-eye is 0.3, and the probability of hitting the nine is 0.4. We need to find the probability that the shooter will have at least 19 points with two shots.
You can't get more than 10 points in one shot and no more than 20 in two shots. It means either he hit the nine with the first shot and the ten with the second, or vice versa. Or he scored 20 points and hit the ten both times.
Let us find the probability. Event A: the shooter scored at least 19 points. Event B1: the shooter made the bull’s eye with the first shot. Accordingly, event B2: the second shot is in the bull’s eye.
Event C1: the first shot is in the nine, event C2: the second shot is in the ten. Then we can represent event A as B1B2 plus B1C2 plus B2C1.
We can calculate the probability of event A as the probability of the sum of all three events.
Since all three events are incompatible, the probability of the sum is equal to the sum of the probabilities. We can write down the probability B1B2 plus the probability B1C2 plus the probability B2C1. Since the shots are independent, the probability of each product can be represented as the product of probabilities.
It is the probability B1 multiplied by the probability B2 plus the probability B1 multiplied by the probability C2 plus the probability B2 multiplied by the probability C2. The probability of hitting the ten is 0.3; the probability of hitting the nine is 0.4.
That is, the probability of events B1 and B2 is 0.3. The probabilities of events C1 and C2 are 0.4. We have 0,3*0,3+0,3*0,4+0,4*0,3=0,33. This is the desired probability. You can start solving problems.