Part 1. Probability theory môn Xác suất thống kê| Trường Đại học Ngoại Thương

Part 1. Probability theory
Chapter 1. Events and their probability
1.1 The concept of trial and event
1.1.1 Definition. A trial is understood as doing an experiment to see if a certain phenomenon occurs or not.
The phenomenon can occur in the result of that trial is called an event.
Example 1. Flip a coin: this is a trial.
The events of this trial are: "Getting tails" and “Getting heads".
Example 2. A box contains 7 quality products and 3 defective products.
Randomly select 2 products: this is a trial.
The events of this trial are: “Getting 2 quality products”; “Getting 2 defective
products”; and “Getting 1 quality product and 1 defective product”.
1.1.2 Classification of events
a) Impossible Event: An impossible event is an event that does not occur when the trial is performed.
The impossible event is denoted by V .
b) Certain event: A certain event is an event that is certain to occur when the trial is performed.
The certain event is denoted by U.
c) Random event: A random event is an event that may or may not occur when the trial is performed.
These events are denoted by A, B, C…
Example 3. Roll a dice: It's a trial
V is an event "getting a side of 7 dots". This event cannot occur when the trial is
performed. This event is an impossible event.
U is an event "getting a side with the number of dots which is more than or equal to 1
and less than or equal to 6". This event always happens when we roll a dice. This event is a certain event.
A is an event "getting a side of 2 dots". When we roll a dice, this event may or may not
happen. So, this event is a random event.
Lecturer: Nguyen Duong Nguyen 1
B is an event "getting a side which has even number of dots". This event is a random event.
1.2 Relationships between events 1.2.1 Sum of events
1) Definition. The sum of two events A and B is the event that occurs if and only if at least one
of the two events A and B occur (either A or B or both occur). The sum of two events A and B is denoted as A + B or AB (the union of A and B).
Similarly, the sum of n events 1
A2,A ,..., nA is the event that occurs if and only if at least one of n the A A k occur. The sum of n events 1 A2,A ,..., nA is denoted as i or . i1
Example 4. Roll a dice. Let A be the event "getting a side of 5 dots", B is the event "getting a
side of 6 dots", then A + B is the event "getting a side with the number of dots which is more than or equal to 5".
Example 5. Three gunners shoot at the same target in turn. Let Ai be the event "The ith gunner
hits the target", i = 1,2,3. Then A1 + A2 + A3 = "At least one gunner hits the target". 2) Properties
Commutative property: A + B = B + A
Combined property: (A + B) + C = A + (B + C) A + A = A; A + U = U; A + V = A 1.2.2 Product of events
1) Definition. The product of two events A and B is the event that occurs if and only if both A
and B occur. The product of two events A and B is denoted as A.B or AB (the intersection of A and B).
Similarly, the product of n events A 1
A2,A ,..., nA is the event that occurs if and only if all the k n
occur. The product of n events A,A ,...,A A 12 n is denoted as i or . i1
Example 6. With A and B being the two events as in Example 4, AB = V.
Example 7. With A1, A2, A3 being the 3 events as in Example 5, A1A2A3 = “All three gunners
hit the target”; A1A2 + A2A3 + A3A1 = "At least two gunners hit the target".
Lecturer: Nguyen Duong Nguyen 2 2) Properties Commutative property: AB = BA
Combined property: A(BC) = (AB)C Distributive property: A(B + C) = AB + AC A + (BC) = (A + B)(A + C) A.A = A; A.U = A; A.V = V
1.2.3 The mutually exclusive events
Definition 1. If A and B are such that A.B = V, they are said to be disjoint or mutually
exclusive (this means that A and B cannot both occur simultaneously in the trial).
Example 8. The two events A and B in Example 4 cannot occur simultaneously, so the two
events are mutually exclusive.
Example 9. Two gunners shoot at the same target in turn. Let A be the event "The first gunner
hits the target", B be the event "The second gunner hits the target". Two events A and B can
occur simultaneously. So, they are not mutually exclusive.
Definition 2. n events A , A , , A are said to be pairwise mutually exclusive if any two of 12 n
these n events are also mutually exclusive.
A , A , , A are pairwise mutually exclusive AiAj = V (i ≠ j; i, j = 1, 2, …, n). 12 n
1.2.4 The complement of an event
1) Definition. The complement of the event A is the event, written A , that occurs when A does not occur.
Remark. If A is the complement of the event A then
i) A and A are mutually exclusive. ii) A+A =U .
Example 10. A box of products contains 7 quality products and 3 defective products.
Randomly pick 2 products from the box. Let A be the event “Get at least 1 quality product”,
then A is the event “Both products obtained are defective products”. 2) Properties +) A B A.B ; A B+C A.B.C . +) AB A B ; ABC A B C . +) AA .
Lecturer: Nguyen Duong Nguyen 3 1.3 Probability of an event
Definition. The probability of event A, written P(A), is a number that measures the ability of
the occurence of the event A when a trial is performed.
1.3.1 Classical definition of probability
Example. Roll a symmetrical and homogeneous dice (fair dice). We see that there are 6
possible cases: getting a side of 1 dot, getting a side of 2 dots, ..., getting a side of 6 dots.
These cases satisfy two conditions: Firstly, they are unique, which means in the trial outcomes,
one and only one of these cases occurs. Secondly, these cases are equally likely to occur. The
cases satisfy the above two conditions are called possible outcomes.
Let A be the event "getting a side which has even number of dots". Out of 6 possible
outcomes, we see only 3 outcomes that, if it occurs then the event A will occur, that is, getting
a side of 2 dots, getting a side of 4 dots and getting a side of 6 dots. These outcomes are called
favorable outcomes to the event A.
Definiton. The probability of the event A in a trial is the ratio between the number of
favorable outcomes to the event A and the number of possible outcomes when performing that trial. P(A) the number of favora ble outcom es to the event A . the num ber of possibl e outc omes
Example. Roll a symmetrical and homogeneous dice. Let A be the event "getting a side which
has even number of dots". Then, P(A) = 3/6 = 0.5.
1.3.2 Statistical definition of probability
Suppose a trial is repeated n times under the same conditions. If the event A occurs k times
then the ratio k/n is called the frequency of occurrence of event A in n given trials, written f(A): f(A ) = k/n Note. 0 f(A) 1.
Example. To study the probability of getting tails when flipping a coin, Buffon and Pearson
flipped a coin several times and got the following result: Frequency Number of flips The number of Experimenters k (n) getting tails (k) f (A) n Buffon 4040 2048 0.5069
Lecturer: Nguyen Duong Nguyen 4 Pearson 12000 6019 0.5016 Pearson 24000 12012 0.5005
When the number of trials changes (n changes), the frequency of the occurrence of
event A also changes, but it always fluctuates around some fixed number. When n is larger, the
frequency of occurrence of event A is closer to that fixed number. That fixed number is called
the probability of event A in the statistical sense.
So, the probability of getting tails when flipping a coin is 0.5.
Note. In fact, when n is large enough, we can approximate P(A) by f(A): P(A) f (A) . 1.3.3 Properties i) 0 P(A) 1. ii P ) U ()1 . iii P) V () 0 .
Example 1. Roll two symmetrical and homogeneous dice simultaneously. Find the probability of getting a sum of 7.
Solution. The possible outcomes when rolling two dice are shown in the following table I II 1 2 3 4 5 6 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Let A be the event "getting a sum of 7".
The number of possible outcomes when the trial is performed is 36.
The number of favorable outcomes to A is 6. 61 So, P(A) = . 36 6
Example 2. Randomly draw 4 cards from a standard 52-card deck. Find the probability that
out of the 4 cards drawn we have:
a) 1 red card and 3 black cards.
b) 2 Hearts, 1 Diamonds, and 1 Spades. c) 2 Aces.
Lecturer: Nguyen Duong Nguyen 5
Solution. The trial is to draw randomly 4 cards. The number of possible outcomes is 4 52! 52.51.50.49 C 270725 . 52 4!48! 24
a) Let A be the event "out of the 4 cards drawn including 1 red card and 3 black cards". 26!
The number of favorable outcomes to A is: 13 C 26 . 26 C 26. 67600 3!23! 67600 P(A) 0.2497 . 270725
b) Let B be the event "out of the 4 cards drawn including 2 Hearts, 1 Diamonds, and 1 Spades".
The number of favorable outcomes to B is: 2 1 1 13! C .C .C .13.13 13182 13 13 13 2!11! 13182 0.0487 P B72 . 270 5
c) Let C be the event "out of the 4 cards drawn including 2 Aces". 4! 48!
The number of favorable outcomes to C is: 22 C 4.48 C . 6768 2!2! 2!46! 6768 0.025 P C7 . 270 25
Example 3. The phone number in the city X is a 7-digit number starting with number 8.
Suppose we randomly select a phone number of that city. Find the probability of selecting a phone number:
a) in which there are 6 remaining digits that are different. b) which is an even number.
c) in which there are 7 different digits, the last number is even.
Solution. We see that the remaining 6 digits of the phone number of the city X are taken from
the set of 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 that the digits of these 6 digits can be the same. 66
Therefore, the number of possible outcomes is A1010 .
a) Let A be the event "Choose a phone number in which there are 6 remaining digits that are
different". The number of favorable outcomes to A is 61 A 0 . So 6 A10 10.9.8.7.6.5 P(A) 0.1512 66 10 10
Lecturer: Nguyen Duong Nguyen 6
b) Let B be the event “Choose a phone number which is an even number”. A phone number is
even if its last digit can be one of the 5 digits 0, 2, 4, 6, 8. The digits of the remaining 5 digits
of the phone number can be the same. Therefore, the number of favorable outcomes to B is: 55 1 A 0.5 10 .5 . We have 5 10 .5 P(B) 0.5 6 10
c) Let C be the event "Choose a phone number in which there are 7 different digits, the last
number is even". Since the first digit of that phone number is 8, the last digit of that phone
number can be 1 out of the 4 digits 0, 2, 4, 6. The remaining 5 digits of that phone number are
different, different from 8 and different from the last digit. Therefore, the number of favorable outcomes to C is 58
A .4 8.7.6.5.4.4 26880 . We obtain that 26880 P(C) 0.2688 6 10 1.4 Multiplication rule
Definition 1. Two events A and B are said to be independent if the occurrence or non-
occurrence of one does not affect the occurrence or non-occurrence of the other event.
Note. If A and B are independent then: A and B are also independent A and B are also independent A and B are also independent
Definition 2. Three events A, A , A are said to be completely independent if each event is 1 2 3
independent of any combination of the other events, which means A and A are independent 12 A23 and A are independent A and A are independent 13 A and A A are independent 1 2 3 A2 and 1 A 3A are independent A and A A are independent 3 1 2
Definition 3. The conditional probability of the event A given the event B, written P(A/B), is
the probability of A computed when knowing that B has occurred.
Lecturer: Nguyen Duong Nguyen 7
Example 1. A box contains 7 quality products and 3 defective products. Randomly pick 2
products from the box. Find the probability that the second product is a defective product,
knowing that the first product is a defective product.
Solution. Let A be the event "the first product is a defective product", B be the event "the
second product is a defective product". The event B occurs after A has occurred. So, the
probability that the second product is a defective product, knowing that that the first product is a defective product is: 2 P(B / A) 0.2222 9 Properties. i) 0 P(A/B) 1 ii) P(A/A) = 1
iii) If the events A and B are independent then P(A/B) = P(A), P(B/A) = P(B).
Theorem 1. The probability of the product of two events A and B is equal to the product of the
probability of one of the two events and the conditional probability of the other event:
P(AB) = P(A).P(B/A) = P(B).P(A/B)
Especially: If the events A and B are independent then P(A.B) = P(A).P(B)
Example 2. A box contains 7 quality products and 3 defective products.
a) Randomly draw 2 products one by one from the box to check. Knowing that after each
inspection, the product is returned to the box. Find the probability of getting 2 defective products.
b) Randomly draw 2 products one by one from the box to check. Know that after each
inspection, the product is not returned to the box. Find the probability of getting 2 defective products.
Solution. Let Ai = “Getting the defective product at the ith time” (i = 1, 2), A = “Getting two
products that are both defective products”. We have: A . A A 12
a) In this case A1, A2 are independent, so we have 33
P(A) = P(A1.A2) = P(A1).P(A2) = . = 0.09 10 10
b) In this case A1, A2 are not independent, so we have
Lecturer: Nguyen Duong Nguyen 8 3 2 1
P(A) = P(A1.A2) = P(A1).P(A2/A1) = . 0.0667 10 9 15 P(AB) Corollary. P(A / B) P(B) .
Example 3. Let P(B) = 1/4; P(AB) 1/ 6 , C is any event. Compute P(A.C A) / B) . Solution. We have P(A.BC AB) P(AB) 1/ 6 2 P(A.C A) / B) P(B) P(B) 1/ 4 3. Theorem 2. Le A t A 1 A be three events. Then 2 ,, 3 P A .A A P A .P A /A P A / A . A 1 23 1 2 1 3 1 2 Especially: Le A t A , A ,
be completely independent events. Then, the probability of the 1 2 3 product of 1 A2,A and 3
A is equal to the product of the component probabilities: P A .A A P A .P A P A 1 2 3 1 2 3
Example 4. A box contains 9 products. For each quality check, 3 products are randomly
selected. After the check is complete, return them to the box. Find the probability that after 3
checks, all products are inspected.
Solution. Let Ai be the event "geting 3 new products on ith check", i = 1,2,3. We see that the
events A1, A2, A3 are dependent on each other. Let A be the event “After 3 times checking that
box, all products are checked”. We have A = A1.A2.A3. 33 CC 5 1 5 P(A) = P(A 63 1)P(A2/A1).P(A3/A1A2) = 1. . 1. . 33 CC 21 84 1764 99 1.5 Addition rule
Theorem 1. The probability of the sum of two events is equal to the sum of the probabilities of
the two events minus the probability of the product of the two events: P(A B) P(A) P(B) - P(A.B) Especially:
+) If A and B are two mutually exclusive events then we have P(A + B) = P(A) + P(B)
+) If A is the complement of the event A then P(A) P(A) 1
Lecturer: Nguyen Duong Nguyen 9
Example 1. Two people shoot at the same target independently. Each person shoots 1 bullet
with the probability of hitting each person is 0.8 and 0.9. Find the probability that at least one person hits the target.
Solution. Let A i = “the ith person hit the target”, i = 1, 2, A = “at least one person hits the
target”. We have A = A1 + A2.
Method 1: Because A1 and A2 are not mutually exclusive,
P(A) = P(A1 + A2) = P(A1) + P(A2) – P(A1.A2)
On the other hand, A1 and A2 are independent. Thus P(A1A2) = P(A1)P(A2) We have
P(A) = P(A1) + P(A2) – P(A1).P(A2) = 0.8 + 0.9 – 0.8.0.9 = 0.98
Method 2: We have A = “Nobody hit the target” and A A12.A
Since A1 and A2 are independent, 1
A and A2 are also independent. So
P(A) P(A 12)P(A ) (1 0.8)(1 0.9) 0.02 P(A) 1 P(A) 1 0.02 0.98
Example 2. A batch of goods (consignment) contains 100 products, including 5 defective
products. That batch is accepted if out of 50 randomly selected product for inspection, the
number of defective products does not exceed 1. Find the probability that the batch of goods is accepted.
Solution. Let A be the event “The batch of goods is accepted”, Ai is the event “Out of 50
selected products, there is i defective product”, (i = 0, 1). We have A = A0 + A1. Because A0
and A1 are mutually exclusive, P(A) = P(A0) + P(A1) We have 50 C95 P(A ) 0.0281 050 C100 1 49 C .C 5 95 P(A ) 0.1529 150 C100 P(A) 0.0281 0.1529 0.181 Corollary:
Lecturer: Nguyen Duong Nguyen 10
i) P((A + C)/B) = P(A/B) + P(C/B) – P((AC)/B))
Especially: If AC = V then P((A + C)/B) = P(A/B) + P(C/B) ii) P(A/B ) = 1 – P(A/B) Theorem 2. Le A t A , A, be events. Then 1 2 3 P A A A ( P A P)A ( P ) ( A )P ( A A ) ( P ) A ( A ) P (A A ) P (A A A ) 1 2 3 1 2 3 1 2 2 3 3 1 1 2 3 Especially: If the eve Ants A 1 2,
A, 3 are pairwise mutually exclusive then P A ( A A P ) A (P ) A ( )P (A ) 1 2 3 1 2 3
Example 3. A train consisting of 3 carriages parks at the platform. There are 5 passengers
stepping on the train. Each passenger independently selects a carriage at random. Calculate the
probability that there is at least one carriage without new passengers boarding.
Solution. Let A = "there is at least one carriage without new passengers boarding ", Ai = "the
ith carriage has no new passengers boarding" (i = 1, 2, 3). We have A = A1 + A2 + A3.
Because A1, A2, A3 are not mutually exclusive,
P(A1+A2+A3) = P(A1) + P(A2) + P(A3) – P(A1A2) – P(A1A3) – P(A2A3) + P(A1A2A3) 5 2 P 1 (A ) ( P ) ( A ) P 3 A ; P (A A ) ( P A) ( A )P 3 A A ; P ( A A A ) 0 1 2 3 5 1 2 2 3 1 3 5 1 2 3 55 P A A A 2 1 31 ( ) 3 3 0 1 2 3 3 3 81 1.6 Bernoulli's formula 1.6.1 Definition
Definition 1. Trials are said to be independent if the results of one do not affect the results of the others.
Definition 2. Suppose we make n independent trials. These n trials are said to be n Bernoulli
trias if it satisfies the following two conditions:
1. In each trial, one of two cases occurs: either the event A occurs or the event A does not occur.
2. The probability of occurring the event A in each trial is the same and equal to p. Therefore,
the probability that the event A does not occur in each trial is equal to q = 1 – p. Example.
+) Flip a balanced and homogeneous coin 10 times. Let A be the event "getting tails", then A be the event " h
getting eads". In each trial, we have
Lecturer: Nguyen Duong Nguyen 11 1 1 p P(A) 2 and q P(A) 2 . Here are 10 Bernoulli trials.
+) Roll a symmetric and homogeneous dice 50 times. Let A = "getting a side of 5 dots", then
A= "getting a side which has the number of dots other than 5". For each trial, we have p = p(A) = 1/6 and q = = 5/6. Here are 50 Bernoulli trials.
+) A box has 6 white balls and 4 black balls. From the box, take out 4 balls one by one
according to the returnable method. Let A be the event "Getting a white ball". The probability
of occurring the event A in each trial is equal to p(A) = 6/10 = 0.6. Here are four Bernoulli
trials. But taking out 4 balls one by one according to the non-returnable method. These are not 4 Bernoulli trials.
1.6.2 Theorem. The probability that in n Bernoulli trials the event A occurs exactly k times k ( n 0 ) , written Pn(k), is P ( ) k k n k k k C p q C p (1 k p ) n k n n n
Example. According to the results of the tuberculosis survey, the rate of people suffering from
tuberculosis in an area is 0.1%. Find the probability that when 10 people are examined:
a) Two people have tuberculosis.
b) At least one person has tuberculosis.
c) At least 9 people have tuberculosis.
Solution. Examining 10 people is 10 Bernoulli trials with the event A = “meeting someone
with tuberculosis” and p(A) = p =0.001. a) P10(2) = 2 2 8 C .(0.001) .(0.999) 10
b) Let B = “at least one person has tuberculosis”. Then B is the event "nobody has tuberculosis". P(B ) = P 0 0 10 10 10(0) = C (0,001) .(0, 999) (0,999) 10
P10(k 1) = 1 – P(B ) = 1 – 10 (0,999) .
c) Let C = “There are at least 9 people with tuberculosis”, D = “There are 9 people with
tuberculosis”, E = “There are 10 people with tuberculosis”.
We have C = D + E. Since D and E are mutually exclusive, 9 9 10 10 0
p(C) P(D) P(E) P (9) P (10) C (0,001) (0,999) C (0, 001) (0,999) 10 10 10 10
Lecturer: Nguyen Duong Nguyen 12