Exercises for Homework 4
4.9. There are details in two boxes: in the first – 10 (3 of them are standard), in the second – 15 (6 of them are standard). One takes out at random on one detail from each box. Find the probability that both details will be standard.
The answer: 0,12.
4.10. There are 3 television cameras in a TV studio. For each camera the probability that it is turned on at present, is equal to p = 0,6. Find the probability that at least one camera is turned on at present (the event А).
The answer: 0,936.
4.11. What is the probability that at least one of a pair of dice lands on 6, given that the sum of the dice is 8?
The answer: 0,4.
4.12. 10 of 20 savings banks are located behind a city boundary. 5 savings banks are randomly selected for an inspection. What is the probability that among the selected banks appears inside the city:
a) 3 savings banks; b) at least one?
The answer: a) 0,348; b) 0,984.
4.13. There are 16 details made by the factory № 1 and 4 details of the factory № 2 at a collector. Two details are randomly taken. Find the probability that at least one of them has been made by the factory № 1.
The answer: 92/95.
4.14. Three buyers went in a shop. The probability that each buyer makes purchases is equal to 0,3. Find the probability that:
a) two of them will make purchases;
b) all three will make purchases;
c) only one of them will make purchases.
The answer: a) 0,189; b) 0,027; c) 0,441.
4.15. Three students pass an exam. The probability that the exam will be passed on "excellent" by the first student is equal to 0,7; by the second – 0,6; and by the third – 0,2. What is the probability that the exam will be passed on "excellent" by:
a) only one student; b) two students;
c) at least one; d) neither of the students?
The answer: a) 0,392; b) 0,428; c) 0,904; d) 0,096.
4.16. Three shots are made in a target. The probability of hit at each shot is equal to 0,6. Find the probability that only one hit will be in result of these shots.
The answer: 0,288.
L E C T U R E 5
Theorem of addition of probabilities of compatible events
Two events are compatible if appearance of one of them doesn’t exclude appearance of another event at the same trial.
Example. A – appearance of four aces at tossing a die; B – appearance of an even number of aces. The events A and B are compatible.
Let events A and B be compatible, and the probabilities of these events and the probability of their joint appearance be given. How can we find the probability of the event A + B consisting in that at least one of the events A and B will appear?
Theorem. The probability of appearance of at least one of two compatible events is equal to the sum of the probabilities of these events without the probability of their joint appearance:
P(A + B) = P(A) + P(B) – P(AB)
Remark 1. Using the obtained formula one should remember that the events A and B can be both independent and dependent.
For independent events: P(A + B) = P(A) + P(B) – P(A) P(B)
For dependent events: P(A + B) = P(A) + P(B) – P(A) PA(B)
Remark 2. If the events A and B are incompatible then their joint appearance is an impossible event and consequently, P(AB) = 0. Then for incompatible events A and B, P(A + B) = P(A) + P(B).
Example. The probabilities of hit in a target at shooting by the first and the second guns are equal respectively: p1 = 0,7; p2 = 0,8. Find the probability of hit at one shot (by two guns) by at least one of guns.
Solution: The probability of hit in the target by each of guns doesn’t depend on result of shooting by another gun, therefore the events A (hit by the first fun) and B (hit by the second gun) are independent. The probability of the event AB (both the first and the second guns gave hit) P(AB) = P(A) P(B) = 0,7 0,8 = 0,56.
The required probability is:
P(A + B) = P(A) + P(B) – P(AB) = 0,7 + 0,8 – 0,56 = 0,94.
Remark
3.
Since in this example the events A
and B
are independent, we can use the formula
(the probability of appearance of at least one of the events). In
fact, the probabilities of the events which are opposite to the
events A
and B,
i.e. the probabilities of misses are:
The required probability that at least one of guns gives hit at one shot is equal to
