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COUNTING TECHNIQUES AND
PROBABILITY - Supplementary Notes (PART 1)
- A menu offers a choice of 3 salads, 8 main dishes, and 5 desserts. How many dif- ferent meals consisting of one salad, one main dish, and one dessert are possible? Solution: Use multiplication rule:
3 × 8 × 5 = 120
- How many three letter “words” can be made from the letters a, b, and c with no letters repeating? A “word” is just an ordered group of letters. It doesn’t have to be a real word in a dictionary.
Solution we use permutations n = 3, r = 3
3 P
3 =^
= 3! = 3 × 2 × 1 = 6
Recall: 0! = 1
- If a 22-member club needs to elect a chair and a treasurer, how many ways can these two to be elected?
Solution we use multiplication rule
22 × 21 = 462
Alternatively we can use permutations as these people will be filling specific positions n = 22, r = 2
(^22) P 2 = 22! (22 − 2)!
22 × 21 × 20!
= 22 × 21 = 462
- A test consist of 12 multiple choice questions, with each question having four pos- sible answers. In how many ways can a student check off one answer to each question?
Solution Each question has 4 possible choices or ways to answer it, we use multi- plication rule
1
4 × 4 × 4 × 4 × 4 × 4 × × 4 × 4 × 4 × 4 × 4 × 4 = 4^12
- A bettor at a racetrack wants to place a single P220 win bet on each of the nine races. If there are 8 horses in each of the first six races and 7 horses in each of the final 3 races, in how many ways can this bettor place his bets? Solution use multiplication rule. note that in each of first 6 races he has 8 horse to choose from and each of the last 3 he has 7 choices.
8 × 8 × 8 × 8 × 8 × 8 × 7 × 7 × 7 = 8^6 × 73
- A committee of 12 is to be split into 3 subcommittees, having 3, 4 and 5 members, respectively. In how many ways can the subcommittees be formed?
Solution We use combination and multiplication rule. The order doesnt matter and the process is done 3 times
12 C 3 × 9 C 4 × 5 C 5 = 12 C 4 × 8 C 5 × 3 C 3 = 12 C 5 × 7 C 3 × 4 C 4
- How many different 4 letter arrangements can we make of the letters in the word ‘cats’, using each letter once only.
Solution: each letter can be use once, means no repetition. The are 4 spaces to be filled by 4 letters. First spot can be filled with any of the 4 letter. In the second spot since we have used one letter and it can no longer be used we have 3 choices. on the third spot since we have use two letter and we cannot used them, we now have 2 choice and the for the last spot we have 1 choice. using multiplication since we are arranging all 4 letters
4 × 3 × 2 × 1 = 4! = 24
alternatively using permutations; n = 4 and r = 4
(^4) P 4 =^
- How many 3 letter ‘words’ can be made using the letters a, b, c, d, e, and f if each letter can be used at most once? Solution: There are 6 letter and we are only arranging 3, no repetition
(^6) P 3 =^
6 × 5 × 4 × 3!
= 6 × 5 × 4 = 120
Alternative use multiplication rule as in Question 7
6 × 5 × 4 = 120
- A quiz consists of 6 multiple choice questions, each with five choices. In how many ways can a student mark the answers to the question if one choice is made for each of the question?
Solution This is similar to Question 4
5 × 5 × 5 × 5 × 5 × 5 = 5^6 = 15625
- How many different letter arrangements can be made from the letters in the word STATISTICS? Solution Similar to Question 11
Let’s break down the word ”STATISTICS”: There are 3 occurrences of the letter ”S.” n 1 = 3 There are 3 occurrences of the letter ”T.”n 2 = 3 There are 2 occurrences of the letter ”I.” n 3 = 2 There is 1 occurrence of the letter ”A.”n 4 = 1 There is 1 occurrence of the letter ”C.”n 5 = 1
Using the formula for permutations with repetition, we calculate the total number of arrangements as: n! n 1 !n 2 !n 3 !n 4 !n 5!
remember n 1 + n 2 + n 3 + n 4 + n 5 = n
- In how many, different ways can a true-false test consisting of 9 questions be an- swered? Solution This is similar to Question 4 and Question 14, but in this case we have 9 questions that can be answered in 2 ways each
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^9
- How many automobile license plates can be made, if each plate contains two dif- ferent letters followed by three different digits?
Solution The first 2 spots are occupied by alphabets (letters) and we have 26 letters The next 3 spots are occupied by digits and we have 10 digits (0 to 10)
2 different letters can be done in 26 × 25
3 digits can be done in 10 × 9 × 8
Then combining the letters and digits for the licence plate is given by
26 × 25 × 10 × 9 × 8
we have used the multiplication rule.
Alternatively using permutations we have (^26) P 2 × (^10) P 3
- A witness to a hit-and-run accident told the police that the license number con- tained the letters RLH followed by 3 digits, the first of which was a 5. If the witness cannot recall the last 2 digits, but is certain that all 3 digits are different, find the maximum number of automobile registrations that the police may have to check.
Solution a. We already know the first 3 letter RLH, no need to worry about that part of the license plate
b. We already know the license plate has 3 different digits and the first one is 5, which means we focus on the last 2 digits and there are 9 digits to choose from (0,1,2,3,4,6,7,8,9). using multiplication rule we get
9 × 8 = 72
Alternatively we can use permutations
(^9) P 2 =^
9 × 8 × 7!
= 9 × 8 = 72
- Construct a sample space that consists of tossing an unbiased die. What is the probability you would roll either a 1, 2, 3, 4, 5, or 6?
Solution
Ω = S = { 1 , 2 , 3 , 4 , 5 , 6 }
each of the outcomes are equally likely
P (1) = P (2) = P (3) = P (4) = P (5) = P (6) =
we are using the word “OR”, so its a union of these events and we know they have no intersections, we sum up their individual probabilities
P (1) + P (2) + P (3) + P (4) + P (5) + P (6) =
- Consider flipping a fair coin twice. Let A be the event where a single head is observed. What is P(A)?
Solution Ω = S = {HH, HT, T H, T T } A = {HT, T H}
P (A) =
number of objects in A number of objects in Ω
- Some trees in a forest were showing signs of disease. A random sample of 200 trees of various sizes was examined yielding the following results:
Type Disease free Doubtful Diseased Total Large 35 18 15 68 Medium 46 32 14 92 Small 24 8 8 40 Total 105 58 37 200
(a) What is the probability that one tree selected at random is large?
P (large tree) =
(b) What is the probability that one tree selected at random is diseased?
P (diseased tree)
(c) What is the probability that one tree selected at random is both small and diseased? P (small and diseased) =
(d) What is the probability that one tree selected at random is either small or disease-free?
P (small or disease free) = P (small) + P (disease free) − P (small and disease free)
=
(e) What is the probability that one tree selected at random from the population of medium trees is doubtful of disease?
P (doubtful given medium) =
medium & doubtful medium
- Suppose you draw one card at random from a standard deck of cards. Recall that a standard deck of cards contains 13 face values (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King) in 4 different suits (Clubs, Diamonds, Hearts, and Spades) for a total of 52 cards. Assume the cards were manufactured to ensure that each outcome is equally likely with a probability of 1/52. Let A be the event that the card drawn is a 2, 3, or 7. Let B be the event that the card is a 2 of hearts (H), 3 of diamonds (D), 8 of spades (S) or king of clubs (C).
Solution Ω = {All the 52 cards } A = { 2 D, 2 C, 2 H, 2 S, 3 D, 3 C, 3 H, 3 S, 7 D, 7 C, 7 H, 7 S} B = { 2 H, 3 D, 8 S, KC}
(a) What is the probability that event A occurs?
P (A) =
(b) What is the probability that event B occurs?
P (B) =
(c) What is the probability that either event A or B occurs?
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
(d) What is P (A ∩ B)? A ∩ B = { 2 H, 3 D}
P (A ∩ B) =
- Given A = { 3 , 4 }, B = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } and C = { 6 , 7 , 8 }
(a) Depict events A, B, and C in a Venn diagram.
A
B
C
(b) Describe the relationship of events A,B, and C.
Solution A is a subset B, B and C have 7 as their common elements (c) List the outcomes comprising the events and Also express each of these events in words i. B′^ = { 8 } All elements in the sample space that are outside set B
ii. A ∩ B = A = { 3 , 4 } All elements in the sample space that are common to set A and B (appear in both sets A and B)
(g)
(B ∩ C)′^ = B′^ ∪ C′^ = that either have an annual turnover not exceeding P10 million or whose financial year does not end in June.
(h) (B ∩ C)′^ ∪ (C ∩ D) To find (B ∩ C)′^ ∪ (C ∩ D), we need to first find the complement of the inter- section of sets B and C and the intersection of sets C and D. Then, we take the union of these two sets.
(B ∩ C)′^ = B′^ ∪ C′= companies that either have an annual turnover not ex- ceeding P10 million or whose financial year does not end in June.
(C ∩ D) companies that have financial year ending in June and their share price is higher now than 6 months ago.
Finally, (B ∩C)′^ ∪(C ∩D)= companies that either have an annual turnover not exceeding P10 million or whose financial year does not end in June, or whose share price is higher now than 6 months ago, or satisfy both conditions.
- In a psychiatric study reported in the New England Journal of Medicine, psychia- trists reported on mental status evaluations.
Letting A be the event that the patients have auditory delusions and
V be the event that the patients have visual delusions. State in words the probabilities expressed by;
(a) P (A) = probability that the patient has auditory delusions
(b) P (V ) = probability that patient has visual delusions (c) P (A ∪ V ) = probability that the patient has auditory delusions or visual delu- sions or both.
(d) P (A ∩ V = probability that the patient has both auditory delusions and visual delusions.
