Theorem of addition of probabilities of incompatible events

 

Let events A and B be incompatible and let the probabilities of these events be known. How can we find the probability of A + B?

Theorem. The probability of appearance of any of two incompatible events is equal to the sum of the probabilities of these events:

P(A + B) = P(A) + P(B)

 

Corollary. The probability of appearance of any of several pairwise incompatible events is equal to the sum of the probabilities of these events:

 

P(A₁ + A₂ + …+ A_n) = P(A₁) + P(A₂)+ … + P(A_n).

Example. There are 30 balls in an urn: 10 red, 5 blue and 15 white. Find the probability of appearance of a colour ball.

Solution: An appearance of a colour ball is an appearance of either red or blue ball. The probability of appearance of a red ball (the event A) is equal to P(A) = 10/30 = 1/3. The probability of appearance of a blue ball (the event B) is equal to: P(B) = 5/30 = 1/6. The events A and B are incompatible (an appearance of a ball of one colour excludes an appearance of a ball of other colour), therefore the theorem of addition is applicable. The required probability is:

 

P(A + B) = P(A) + P(B) = 1/3 + 1/6 = 1/2.

Example. A shooter shoots in a target subdivided into three areas. The probability of hit in the first area is 0,45 and in the second – 0,35. Find the probability that the shooter will hit at one shot either in the first area or in the second area.

Solution: The events A – «the shooter hit in the first area» and B – «the shooter hit in the second area» are incompatible (hit in one area excludes hit in other area). Therefore, the theorem of addition is applicable. The required probability is:

P(A + B) = P(A) + P(B) = 0,45 + 0,35 = 0,80.

 

Complete group of events

Theorem. The sum of the probabilities of events A₁, A₂, A₃, …, A_n which form a complete group is equal to 1:

P(A₁) + P(A₂) + P(A₃) + … + P(A_n) = 1

Example. A consulting point of an institute receives packages with control works from the cities A, B and С. The probability of receiving a package from the city A is equal 0,7; from the city B – 0,2. Find the probability that next package will be received from the city С.

Solution: The events «a package has been received from A», «a package has been received from B» and «a package has been received from C» form a complete group. Therefore, the sum of probabilities of these events is equal to 1:

 

0,7 + 0,2 + p = 1

 

Then the required probability is equal to p = 1 – 0,9 = 0,1.

 

Opposite events

Two uniquely possible events forming a complete group are opposite. If A denotes one of two opposite events, then the opposite to A event is denoted by .

Example. Hit and miss at a shot in a target are opposite events.

Example. A detail is randomly taken from a box. The events «a standard detail has appeared» and «a non-standard detail has appeared» are opposite.

Theorem. The sum of the probabilities of opposite events is equal to 1:

P(A) + P() = 1

Example. The probability that a day will be rainy is p = 0,7. Find the probability that a day will be clear.

Solution: The events «a day is rainy» and «a day is clear» are opposite, therefore the required probability is q = 1 – p = 1 – 0,7 = 0,3.

Example. There are n details in a box, and m of them are standard. Find the probability that there is at least one standard detail among k randomly extracted details.

Solution: The events «there is at least one standard detail among the extracted details» and «there is no standard detail among the extracted details» are opposite. Denote the first event by A, and the second – by . Obviously, P(A) = 1 – P (). Find P(). The total number of ways by which one can extract k details from n details is The number of non-standard details is n – m. One can extract k non-standard details from n – m non-standard details by ways. Therefore, the probability that there is no standard detail among k extracted details is The required probability is:

Conditional probability

A random event has been before determined as an event that can take place or not to take place for holding the set of conditions S. If there are no any other restrictions except for the conditions S for a calculation of the probability of an event then such a probability is unconditional; if there are other auxiliary conditions then the probability of an event is said to be conditional. For example, the probability of an event B is very often calculated with an auxiliary condition that an event A was happened. Observe that unconditional probability also, strictly speaking, is conditional since it is supposed that the conditions S hold.

The conditional probability P_A(B) is the probability of the event B calculated in assumption that the event A has already happened.

Example. There are 3 white and 3 black balls in an urn. One takes out twice on one ball from the urn without replacement. Find the probability of appearance of a white ball at the second trial (the event B) if a black ball was extracted at the first trial (the event A).

Solution: There are 5 balls after the first trial, and 3 of them are white. The required probability is P_A(B) = 3/5.

The conditional probability of an event B with the condition that an event A has already happened is equal to:

where P(A) > 0.

 

Indeed, return to our example. The probability of appearance of a white ball at the first trial P(A) = 3/6= 1/2. Find the probability P(AB) that a black ball will be appeared at the first trial and a white ball – at the second trial. The total number of events – a joint appearance of two balls (indifferently from colour) is equal to the number of allocations The event AB is favored 3 × 3 = 9 events from the total number. Consequently, P(AB) = 9/30 = 3/10. Thus,

 

P_A(B) = P(AB)/P(A) = (3/10)/(1/2) = 3/5.