Exercises for Homework 1
1.11. A die is tossed. Find the probability that the upper side of the die shows:
a) six aces; b) an odd number of aces;
c) no less than four aces; d) no more than two aces.
1.12. An urn contains 12 balls: 3 white, 4 black and 5 red. Find the probability that a randomly taken ball will be black.
1.13. The first box contains 5 balls with numbers from 1 up to 5, and the second – 5 balls with numbers from 6 up to 10. It has been randomly extracted on one ball from each box. Find the probability that the sum of numbers of the extracted balls will be: 1) no less than 7; 2) equal to 11; 3) no more than 11.
The answer: 2) 0,2; 3) 0,6.
1.14. The relative frequency of workers of an enterprise having a higher education is equal to 0,18. Determine the number of workers having a higher education if the total number of workers of the enterprise is 350.
1.15. 78 seeds have germinated of 100 planted seeds. Find the relative frequency of germination of seeds (seed – семя; to germinate – прорастать).
1.16. Two dice are tossed. Find the probability that:
a) the sum of aces equals 5, and the product equals 6;
b) the product of aces doesn’t exceed 6;
c) the product of aces is divided on 6.
1.17. A point C is randomly appeared in a segment AB of the length 3. Determine the probability that the distance between C and B exceeds 1.
The answer: 2/3.
1.18. A point is randomly thrown inside of a circle of the radius R. Find the probability that the point will be inside the square entered in the circle. It is supposed that the probability of hit of a point in the square is proportional to its area and does not depend on its location regarding the circle.
The answer: 2/.
1.19. A coin is tossed twice. Find the probability that the coin lands on heads in both times.
The answer: 1/4.
1.20. A point B(x) is randomly put in a segment OA of the length 6 of the numeric axis Ox. Find the probability that both the segments OB and BA have the length which is greater than 2.
The answer: 1/3.
L E C T U R E 2
Basic formulas of combinatorial analysis
Here is a typical problem of interest involving probability. A communication system is to consist of n seemingly identical antennas that are to be lined up in a linear order. The resulting system will then be able to receive all incoming signals – and will be called functional – as long as no two consecutive antennas are defective. If it turns out that exactly m of the n antennas are defective, what is the probability that the resulting system will be functional? For instance, in the special case where n = 4 and m = 2 there are 6 possible system configurations – namely,
0110, 0101, 1010, 0011, 1001, 1100
where 1 means that the antenna is working and 0 that it is defective. As the resulting system will be functional in the first 3 arrangements and not functional in the remaining 3, it seems reasonable to take 3/6 = 1/2 as the desired probability. In the case of general n and m, we could compute the probability that the system is functional in a similar fashion. That is, we could count the number of configurations that result in the system being functional and then divide by the total number of all possible configurations.
From the preceding we see that it would be useful to have an effective method for counting the number of ways that things can occur. In fact, many problems in probability theory can be solved simply by counting the number of different ways that a determinate event can occur. The mathematical theory of counting is formally known as combinatorial analysis.
Combinatorial analysis studies quantities of ordered sets subordinate to determinate conditions, which can be made of elements, indifferent of a nature, of a given finite set.
Permutations are ordered sets consisting of the same n different elements and distinguishing only by the order of their disposition (location).
The number of all possible permutations Pn = n!
where
Observe that it is convenient to consider 0!, assuming by definition that
0
!
= 1
Example. How many three-place numbers can be made of the digits 1, 2, 3 if each digit is included into the image of a number only once?
Solution: P3 = 3! = 1 2 3 = 6.
Allocations are ordered sets composed of n different elements on m elements, which are differed by either structure of elements or their order.
The number of all possible allocations
,
i.e.
Example. How many signals is it possible to make of 6 flags of different colour taken on 2?
Solution:
C
ombinations
are ordered sets composed of n
different elements on m
elements that are differed by at least one element.
The
number of all possible combinations
Example. How many ways are there to choose two details from a box containing 10 details?
Solution:
The required number of ways
Observe
that the numbers of allocations, permutations and combinations are
connected by the equality
R
emark.
It
was above supposed that all n
elements are different. If some elements are repeated then in this
case ordered
sets with repetitions
are calculated by other formulas. For example, if there are
n1
elements of one kind, n2
elements of other kind and et cetera among n
elements then the
number of permutations with repetitions
where n1 + n2 + … = n.
Example. How many seven-place numbers consisting of the digits 4, 5 and 6 are there in which the digit 4 is repeated 3 times, and the digits 5 and 6 – 2 times?
Solution: Each seven-place number differs from other by the order of consecution of the digits (so that n1 = 3, n2 = 2, n3 = 2, and their sum is equal to 7), i.e. is the permutation with repetitions of 7 elements:
If in allocations (combinations) of n elements on m some of elements (or all) can appear identical, such allocations (combinations) are said to be allocations (combinations) with repetitions of n elements on m. For example, allocations with repetitions of 5 elements a, b, c, d, e on 3 are abc, cba, bcd, cdb, bbe, ebb, beb, ddd and et cetera; combinations with repetitions – abc, bcd, bbe, ddd and et cetera.
The number of allocations with repetitions of n elements on m is
The number of combinations with repetitions of n elements on m is
Example. 10 films participate in a competition on 5 nominations. How many variants of distribution of prizes are there, if on each nomination are established:
a) different prizes; b) identical prizes?
Solution: a) Each of variants of distribution of prizes represents a combination of 5 films from 10 that differs from other combinations by both the structure of elements and their order on nominations so that the same films can be repeated some times (a film can receive prizes on both one nomination and some nominations), i.e. it represents the allocations with repetitions of 10 elements on 5:
b) If identical prizes are established on each nomination then the order of following the films in a combination of 5 prizewinners doesn’t have any significance. Therefore, the number of variants of distribution of prizes represents the number of combinations with repetitions of 10 elements on 5:
The following rules are used for solving problems of combinatorial analysis:
Sum rule. If some object A can be chosen from the set of objects by m ways, and another object B can be chosen by n ways, then we can choose either A or B by m + n ways.
Example. There are 300 details in a box. It is known that 150 of them are details of the first kind, 120 – the second kind, and the rest – the third kind. How many ways of extracting a detail of the first or the second kind from the box are there?
Solution: A detail of the first kind can be extracted by n1 = 150 ways, of the second kind – by n2 = 120 ways. By the sum rule there are n1 + n2 = 150 + 120 = 270 ways of extracting a detail of the first or the second kind.
Product rule. If an object A can be chosen from the set of objects by m ways and after every such choice an object B can be chosen by n ways then the pair of the objects (A, B) in this order can be chosen by mn ways.
Example. There are 30 students in a group. It is necessary to choose a leader, its deputy and head of professional committee. How many ways of choosing them are there?
Solution: A leader can be chosen from any of 30 students, its deputy – from any of the rest 29 students, and head of professional committee – from any of the rest 28 students, i.e. n1 = 30, n2 = 29, n3 = 28. By the product rule the total number of ways is n1 n2 n3 = 30 29 28 = 24360.