Counting Principle
Multiplication RuleIf an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two experiments a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in (n1) (n2) (n3)… (nk) ways.
Ex 1 For positive numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 0if each number is not allowed to be used more than once,
- how many 3-digit numbers can be constructed?
- In how many ways can you construct even numbers?
- In how many ways can you construct integers greater than 600?
Sol. (a) 9 x 9 x 8 = 648
(b) ending with 0 = 9 x 8 x 1 = 72
ending with 2, 4, 6, 8 = 8 x 8 x 4 = 256
except 0,and(2, 4, 6, 8)
72 +256 =…..
(c) integers greater than 600= 4 x 9 x 8 = 288
Permutation
N factorial : n! = products of positive integers from 1 ton
n! = n (n-1) (n-2)…3.2.1
0 ! = 1
Permutations using all the objects
A permutation of n objects, arranged into one group of size n, without repetition, and order being important is:nPn = P(n,n) = n!
Example: Find all permutations of the letters "ABC" =3! = 3*2*1
= ABC ACB BAC BCA CAB CBA = 6 ways
Permutations of some of the objects
Theorem : The number of ordered arrangements, or permutation , of r objects selected from n distinct objects (r n) is given by:= n (n –1) . . . (n – r + 1) =
Ex. 2 A public bus has 8 empty seats. If 4 passengers get on the bus, in how many ways can they be seated?
= = 8 x 7 x 6 x 5 = 1,680 ways
Ex. 3 If 8 students areto be queued, find the number of ways in which
(a) Tom and Nid are next to each other
(b) Tom and Nid are at each end of the line.
(c) Tom and Nid are not allowed to stand next to each other
Sol.
Combination
The number of ways in which r objects can be selected from a set of n distinct objects (r n) is:
= = == =
Ex. 4 In playing poker, 5 cards are picked from a deck of cards. Find the number of ways in which
(a) all the five cards are heart
(b) all the five cards are red color
(c) 4 out of 5 are kings
Sol. (a) all the five cards are heart = = 1,287 ways
(b) all the five cards are red color = = 65,870 ways
( c ) 4 out of 5 are kings= = 48 ways
Arrangement of distinct objects in circle
We have to fix one object, then perform permutation of the other n-1. This can be done in (n – 1)! waysEx. 5 In a dance there are 4 men and 4 women. Find the number of ways in which
(a) they are arranged in circle where all men stand next to one another
(b) they are arranged in circle where man and woman stand next to each other
Sol. (a) 2 (3! 4!)
(b) 3! 4!
Partitioning
The number of ways of partitioning n distinct objects into k groups containing n1, n2, . . . , nk objects, respectively, iswhere
Ex.6 Suppose that 10 employees are to be divided among 3 jobs with 3 employees going to job I, 4 to job II, and 3 to job iii. In how many ways can the job arrangement be made?
Sol. = = 4,200
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Conditional Probability
If A and B are any events in sample space S, then the conditional probability A given B, denoted by P (A \ B), is defined by:
P (A \ B)= ; P (B) 0
And conditional probability of B given A is
P (B \ A)= ; P (A) 0
Note P (A \ B) = conditional probability of A given that B has occurred.
Ex. 7 A bin contains 5 defective transistors (that immediately fail when put in use), 10 partially defective (that fail after a couple of hours of use) and 25 acceptable transistors. A transistor is chosen at random from the bin and put into use. If it does not immediately fail, what is the probability it is acceptable?
Sol. Since the transistor did not immediately fail, we know that it is not one of the 5 defectives and so the desired probability is:
P {acceptable\ not defective} =
The transistor will be both acceptable and not defective if it is acceptable, thus
= = 5 / 7
acceptable + partially defective
25 + 10
Ex. 8 If the probability that a communication system will have high fidelity is 0.81 and the probability that it will have high fidelity and high selectivity is 0.18, what is the probability that a system with high fidelity will also have high selectivity?
Sol. Let A : event that a communication system has high selectivity
B: event that a communication system has high fidelity
Given P (A B) = 0.18
P (B) = 0.81
P (A \ B) = = 0.18 / 0.81 = 2 / 9
Ex. 9 If the probability that a research project will be well planned is 0.8 and the probability that it will be well planned and well executed is 0.72, what is the probability that a research project that is well planned will also be well executed?
Sol. Let A : the research project is well executed
B : the research project is well planned
Given P(B) = 0.8,P (A B) = 0.72
P (A \ B) = = 0.72 / 0.8 = 0.9
Ex. 10 Mr. Jones figures that there is a 30 % chance that his company will set up a branch office in Pheonix. If it does, he is 60% certain that he will be made manager of this new operation. What is the probability that Jones will be a Phoenix branch office manager?
Sol. Let B: the company sets up a branch office in Phoenix, P (B) = 0.3
M: Jones is made the manager
P (B M) = P (B) P (M \ B) = (0.3) (0.6) = 0.18
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Rule of Multiplication of Probabilities
If A and B are events in a sample space S, thenP (A B) = P (A \ B) P (B)
= P (B \ A) P (A)
This rule is important because it is often the case that P (A B) is desired, whereas both P (B) and P (A \ B) can be specified from the problem description.
The multiplication rule is most useful when the experiment consists of several stages in succession. If the conditioning event B describes the outcome of the 1st stage, and A the outcome of the second, the P (A\B) = conditioning on what outcome occurs first – will often be known.
This rule is easily extended to experiments involving more than 2 stages. For example,
P (A1 A2 A3) = P (A3 \ A1A2 ) P (A1 A2)
= P (A3 \ A1A2 ) P (A2 \ A1) P (A1)
Ex. 11 There are 2 bags. The first one has 4 white balls and 3 black balls. The second bag contains 3 white balls and 5 black balls. If a ball is randomly chosen from the first bag and put into the second bag, find the probability that a black ball is picked from the second bag.
Sol.
Let B1 = a black ball from 1st bag , B2 = a black ball from 2nd bag
W1 = a white ball from 1st bag , W2 = a white ball from 2nd bag
Independent Events
A andB are independent events if the occurrence of one event does not have any influence on the occurrence of another event.
P (A \ B) = P (A)
P (B \ A) = P (B)
From P (A \B) = , P (B \ A) =
ThereforeP (A B) = P (B) P (A \ B) = P(B) P(A)
P (A B) = P (A) P (B \ A) = P(A) P(B)
P (A B) = P (A) P (B)= P (B) P (A)
Rules of Multiplication of Probabilities
Ex. 12 Four individuals have responded to a request by a blood bank for blood donations. None of them has donated before, so their blood types are unknown. Suppose that only type A+ is desired, and that only one of the four actually has this type. If the potential donors are selected in random order for typing, what is the probability that at least three individuals must be typed to obtain the desired type?
Sol.
Ex. 13 A chain of video stores sells three different brands of videocassette recorder (VCRs). Fifty percent of its VCR sales are brand 1, 30% are brand 2, and 20% are brand3. Each manufacturer offers a one-year warranty. It is known that 25 % of brand 1 require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20 % and 10 % respectively.
(a)What is the probability that a randomly selected purchaser has bought a brand 1 VCR that will need repair while under warranty?
(b)What is the probability that a randomly selected purchaser has a VCR that will need repair while under warranty?
(c)If a customer returns to the store with a VCR that needs warranty repair work, what is the probability that it is a brand 1 VCR? A brand 2 VCR? A Brand 3 VCR?
Sol. Let Ai = {brand i is purchased} for i = 1, 2, 3
B = { needs repair}
Given P (A1) = 0.5, P (A2) = 0.3, P (A3) = 0.2
P (B\ A1) = 0.25, P (B\ A2) = 0.2, P (B\ A3) = 0.1
Wanted: (a) P (A1 B)
(b)P(B)
(c)P (A1 \B), P (A2 \B), P (A3 \B)
Use tree diagram
Independence of more than 2 events
If A, B, C are 3 events, A, B, C are independent if
1. P (A B) = P(A) P(B) and
2. P (B C) = P(B) P(C) and
3. P (C A) = P(C) P (A) and
4. P (A B C) = P (A) P(B) P(C)
Ex.14 Consider an experiment that consists of rolling a single die once and then tossing a fair coin once:
S = { (1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}
Given that these 12 outcome are equally likely, find P (A B) if
A : the die shows 1 or 2
B : the coin shows head
Ex. 15 During a space shot, the primary computer system is backed up by two secondary systems. They operate independently of one another, and each is 90% reliable. What is the probability that all 3 systems will be operable at the time of the launch?
Ex. 16 Consider an analysis of seawater samples taken near the mouth of a river of which numerous industrial plants are located. If A1 is the event that toxic levels of lead are found, and A2 is the event that toxic levels of mercury are detected. Given P(A1 ) = 0.32, P(A2 ) = 0.16, and P (A1 A2 ) = 0.1. Are these events independent?
Ex. 17 Find the probability of getting two heads in 2 flips of a balanced coin.
Ex. 18 A system consists of 4 components as illustrated below. The entire system will work if either the 1-2 subsystem works or if the 3-4 subsystem works (since the two subsystems are connected in parallel). Since the two components in each subsystem are connected in series, a subsystem will work only if both its components work. If components work or fail independently of one another, and if each works with probability 0.9, what is the probability that the entire system will work (the system reliability coefficient)?
Rules of Elimination
If the mutually exclusive events B1 , B2 , . .. , Bnconstitute a partition of the sample space S such that P(Bi) 0 for i = 1, 2, …, n, then for any event A in S
P (A) = =
For mutually exclusive events,
A = (B1 A) (B2 A) . . . (Bn A)
P (A)= P [(B1 A) (B2 A) . . . (Bn A) ]
= P (B1 A) (B2 A) . . . (Bn A)
=
= from Multiplication Rule
Bayes’ Rule
If the mutually exclusive events B1 , B2 , . .. , Bnconstitute a partition of the sample space S such that P(Bi) 0 for i = 1, 2, …, n, then for any event A in S such that P(A) 0
P (Bj \A) = = =
=
P (Bj \A) = Bayes’ RuleBayes’ Rule is an extremely useful formula. It can be used to find the conditional probability P(Bj\A) when the available information cannot be used to apply the definition of conditional probability directly.
Ex. 19 A factory manufactures products by using 4 machines which can produce 1000, 1200, 1800, and 2000 pieces a day, respectively. The machines produce defective products at the rates of 1 %, ½ %, ½ %, and 1 %, respectively.A product is selected from a lot which is produced in one day and found to be defective. Determine the probability that this product is manufactured by the 4th machine.
Sol.
Let B1 , B2 , B3 , B4 be the events that the product is made by machine 1, 2, 3, 4, respectively
D be the event that the product is defective
Ex. 20 The blood type distribution in the USA is type A 41%, Type B 9 %, type AB 4 %, type O 46 %. It is estimated that during World War II 4 % of inductees with type O were typed as having type A; 88 % of those with type A were correctly typed; 4% of type B blood were typed as A; and 10% with type AB were typed as A. A soldier was wounded and brought to surgery. He was typed as having type A blood. What is the probability that this is his true blood type?
Ex. 21 The Gimmick TV Model A uses a printed circuit, and the company has a routine method for diagnosing defects in the circuitry, if a set fails. Over the years, the experience with this method yields the following pertinent information:
- The probability that a set, which fails due to printed circuit defects (PCD) is correctly diagnosed failing because of PCD, is 80%
- The probability that a set, which fails due to causes other than PCD, is diagnosed incorrectly as failing because of PCD, is 30%
Further, experience with the printed circuit shows that about 25% of all model A failures are due to PCD.
Find the probability that a model A set’s failure is due to PCD, given that it has been diagnosed as being due to PCD.