Exercises for Homework 2
2.13. How many different 7-place codes for license plates are possible if the first 3 places are to be occupied by letters of Latin alphabet and the final 4 by numbers?
The answer: 175760000.
2.14. In Ex. 2.13, how many codes for license plates would be possible if repetition among letters or numbers were prohibited?
2.15. 10 persons participate in competitions, and three of them will take the first, second and third places. How many different variants are possible?
The answer: 720.
2.16. How many ways of choosing 3 persons of 10 are possible?
The answer: 120.
2.17. A randomly taken phone number consists of 5 digits. What is the probability that all digits of the phone number are:
a) identical;
b) odd?
It is known that any phone number does not begin with the digit zero.
The answer: a) 0,0001; b) 0,0347.
2.18. There are 3 cards with letter S, 3 cards with letter T, 2 cards with letter I, 1 card with letter A and 1 card with letter C. Cards are mixed and randomly taken out without replacement by one. Find the probability that cards with letters are taken out by the way of consecution of letters of the word «STATISTICS».
The answer: 0,0000198.
2.19. A box contains 15 details, and 10 of them are painted. A collector chooses at random 3 details. Find the probability that the chosen details are painted (collector – сборщик).
The answer: 0,264.
2.20. Find the probability that from 10 books located in a random order, 3 certain books will be beside.
The answer: 0,0667.
2.21. Four tickets are distributed among 25 students (15 of them are girls). Everyone can take only one ticket. What is the probability that owners of these tickets will be:
a) four girls;
b) four young men;
c) three young men and one girl?
The answer: a) 0,108; b) 0,017; c) 0,142.
2.22. There are 100 products (including 4 defective) in a batch. The batch is arbitrarily divided into two equal parts which are sent to two consumers. What is the probability that all defective products will be got:
a) by one consumer;
b) by both consumers fifty-fifty?
The answer: a) 0,117; b) 0,383.
2.23. A library consists of ten different books, and five books cost on 4 thousands of tenghe each, three books – on one thousand of tenghe and two books – on 3 thousands of tenghe. Find the probability that two randomly taken books cost 5 thousands of tenghe.
The answer: 1/3.
2.24. A coin is tossed three times. Let Ai be the event «an appearance of heads at the i-th tossing» (i = 1, 2, 3). Express by A1, A2, A3 and their negations the following events: A – «three heads»; B – «three tails»; C – «at least one heads»; D – «at least one tails»; E – «only one heads»; F – «only one tails».
L E C T U R E 3
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 + …+ An) = P(A1) + P(A2)+ … + P(An).
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.