Elementary Statistics

Transcription

1 Elementary Statistics Chapter 03 Dr. hamsary Page 1 Elementary Statistics M. hamsary, Ph.D. Chapter 03 1

2 Elementary Statistics Chapter 03 Dr. hamsary Page 2 Counting Techniques Multiplication Rule(Principal of Multiplication) Permutations Combinations Multiplication Rule: A. For a sequence of 2 events in which the first one has m possibilities of happening and the second event has n, then together can occur m.n possibilities.. For a sequence of k events in which the first one has n 1, the first one has n 2,, and so on till the kth one has n k possibilities of happening, then together they can occur in n 1. n 2. n 3.. n k possibilities. Example 1: A single coin is flipped and A single die is rolled at the same time. How many different outcomes are possible? Solution: The coin has 2 possible outcomes, the die has 6 possible outcomes. Together there are 2x6 = 12 possible outcomes. (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6). Example 2: Suppose your teacher own 4 pairs of shoes, 5 pairs of pants, 10 shirts and 3 jackets. How many different outfits can he make? ( An outfit consists of one of each items). Solution: y multiplication rule, there are = 600 possible outfits. 2

3 Elementary Statistics Chapter 03 Dr. hamsary Page 3 Example 3: Two coins are tossed together. Find the number of all possible outcomes. Solution: The first coin has 2 possible outcomes, the second has 2 possible outcomes. Together there are 2x2 = 4 possible outcomes. {H H, HT, TH, TT} Example 4: Two dice are rolled together. Find the number of all possible outcomes. Solution: The first die has 6 possible outcomes, the second die has 6 possible outcomes. Together there are 6x6 = 36 possible outcomes.r (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) Example 5: Three coins are tossed together. Find the number of all possible outcomes. Solution: The first coin has 2 possible outcomes, the second has 2 possible outcomes, and so does the third coin. y multiplication rule, there are 2x2x2 = 8 possible outcomes (Not 6). First Second Third H H H H T T T T H H T T H H T T H T H T H T H T Notice the pattern in the columns : In the first column, there is a sequence 4 H and 4 T. In the second column, there is sequence 2H and 2T. In the last column, there is sequence 1H and 1T sequence. 3

4 Elementary Statistics Chapter 03 Dr. hamsary Page 4 Example 6: Three dice are rolled together. Find the number of all possible outcomes. Solution: The first die has 6 possible outcomes, the second die has 6 possible outcomes, and so does the third one. y multiplication rule, there are = 216 possible outcomes (Not 18). It is very tedious to list them all, but let us list a few of them. (1, 1,1), (1,1, 2), (3, 4, 5) (6, 6, 5), (6, 6, 6). Example 7: Tom was given a phone number by his girl friend long time ago. Last Friday night he missed her so much and he tried to call her. He could not find her phone number, but he knew the area code and the first 3 digits (909) He decided to try all phone numbers start with 985. How many phone numbers are in that area code with prefix 985. Solution: First he tried (909) , then (909) , (909) ,.,, and the last possible number is (909) (Off course he does not have to try all of these possibilities). So there are 10,000 different phone numbers start with (909) 985- Now Let us use the multiplication rule. Area Code Prefix The last 4 digits (909) st 2nd 3rd 4th The 1st place on the last digit or suffix can be 0 t0 9 which is 10 possibilities, The 2nd 0 to 9 The 3rd 0 to 9 The 4th place on the last digit or suffix can be 0 t0 9 which is 10 possibilities. y multiplication rule, there are 10x10x10x10 = 10,000 possible outcomes. Example 8: The State of California used to issue license plates consisting of 3 digits, followed by 3 letters such as 245HT. Currently they issue the license plates with 7 positions. 4

5 Elementary Statistics Chapter 03 Dr. hamsary Page 5 They consist of 1 digit (no 0 in the first place), followed by 3 letters, and 3 digits in that order such as 3VK523. Find the total number of license plates in each case. Solution: In order to count all possible numbers, again we apply the multiplication rule as follows: Since there are 10 digits from 0 to 9 and 26 letters A to Z. efore: 10 x 10 x 10 x 26 x 26 x 26 =17,576,000. After: 9 x 26 x 26 x 26 x 10 x 10 x 10 =158,184,000. Example 9: A student is taking a true false test with 10 questions, but he randomly filling out the test without knowing the answer to any of them. How many possibilities are there? Solution: Each question has 2 possible answers, True or False. So by the multiplication rule, there are 2x2x..x2 = 2 10 =1,024 possible filling out. Example 10: Do example 9, if there are 10 multiple choice questions with 5 possible answers for each one. Solution: Each question has 5 possible answers, A,, C, D, and E. So by the multiplication rule, there are 5x5x..x5 = 5 10 =9,765,625 possible filling out. Remark: In general If k coins are tossed, then are 2 K possible outcomes. If k dice are rolled, then are 6 K possible outcomes. If m coins are tossed and n dice are rolled, then are 2 m.6 n possible outcomes. 5

6 Elementary Statistics Chapter 03 Dr. hamsary Page 6 What If Repetition Not Allowed? Example 11: Do example 7, if repetition is not allowed. Solution: y multiplication rule, there are 10x9x8x7 = 5,040 possible numbers. The 1st place on the last digit or suffix can be 0 t0 9 all 10 digits can be used, The 2nd 0 to 9, but we cannot use all 10, only 9 The 3rd 0 to 9, but we cannot use all 10, only 8 The 4th place on the last digit or suffix can be 0 t0 9 that are not used in the previous locations. So we can only use 7 digits. In above example we have no repetition. This kind of counting are called permutation or combination. To understand this concept better, let us try the following example with 3 letters A,, and C. Example 12: iven 3 letters A,, and C, how many 3 letter words can be arranged? Solution: There are 3 cases we have to consider, as shown in the following table. Case I: Repetition allowed: There will be 3*3*3=27 possible arrangements AAA CCC AA C CCA AAC A CC AA C CAC A CC CAA AC CA CA ACA A CC AC AC CA ACC AA C Case II: Repetition not allowed, but order counts: There will be 3*2*1=6 possible arrangements. AC AC AC CA CA CA. Case III: Repetition not allowed and order does not count: There will be only 1 possible arrangement, namely AC 6

7 Elementary Statistics Chapter 03 Dr. hamsary Page 7 Summary of all 3 cases: Repetition Allowed Repetition Not Allowed AAA CCC AA C CCA AAC A CC AA C CAC A CC CAA AC CA CA ACA A CC AC AC CA ACC AA C There are 3x3x3 =27 ways Permutation Order counts AC AC AC CA CA CA There are 3x2x1 =3! = 6 ways Combination Order does not count AC There is only (3x2x1)/3! =1 way Example 13: Now given 4 letters A,, C, and D, and select 2 of these letters to make a word with only 2 letters. Solution: As in example 12, there are 3 cases we have to consider as follows: Case I: Repetition allowed: There will be 4*4=16 possible arrangements AA A CA DA A C D AC C CC DC AD D CD DD Case II: Repetition not allowed, but order counts: There will be 4*3=12 possible arrangements. A A CA DA AC C C D AD D CD DC Case III: Repetition not allowed and order does not count: There will be only 4*3/2=6 possible arrangements, A A CA DA AC C C D AD D CD DC 7

8 Elementary Statistics Chapter 03 Dr. hamsary Page 8 Summary of all 3 cases: Repetition Allowed Repetition Not Allowed AA A CA DA A C D AC C CC DC AD D CD DD There are 4x4 =16 ways Permutation Order counts A A CA DA AC C C D AD D CD DC There are 4x3 =12 ways Combination Order does not count A A CA DA AC C C D AD D CD DC There are (4x3)/2! =6 ways Permutation: A. A collection of n different objects can be arranged in n! ways.. The number arrangement of selecting r items from n objects is called permutation if the order is counted and it is denoted by npr which defined by n Pr = n! n r! ( ). Combination: The number arrangement of selecting r items from n objects is called combination if the order is not counted and it is denoted by ncr which defined by nc r = n!. r!n r! ( ) 8

9 Elementary Statistics Chapter 03 Dr. hamsary Page 9 Example 14: iven 4 letters A,, C, and D, then select 2 of these letter to make a word. Find number of possibilities in the following 2 cases: Order counts Order does not count Solution: In both cases we have n=4 and r=2 Order counts: If order counts, then the problem is permutation rule and we have n! n! npr = ( r ) 4! = = = = 12, 4 2! 2! 2 = 4P2= ( ) Order does not count If order does no count, then the problem is combination rule and we have ncr = n! r! n! ( r) 4! =4C2 = = = = 6 which is the same as the 2! 4 2! 2! 2! 2x2 ( ) number given in the table of example 13. Example 15: In a horse race involving 10 horses, how many ways can first, second, and third places be decided? Solution: This is clearly permutation problem, since the order must be considered. We have n = 10, r =3, So we have 10P3 = n! 10! 10! 10* 9* 8* 7! n = = = = 720 b r g! b10 3g! 7! 7! ways of doing the selection. Example 16: In the poker, a hand consists of 5 cards. How many possible poker hands are there in an ordinary deck of 52 cards? 9

10 Elementary Statistics Chapter 03 Dr. hamsary Page 10 Solution: This is clearly combination problem, since the order of drawing is not important. We have n = 52, r =5, ncr = n r n! 52C5 = = r! ( n r )! ( ) 52 52! 52! = = = = 2, 598, ! 52 5! 5! 47! ways of drawing a hand of 5 cards. Example 17: Do example 16, if we want 3 red cards and 2 black cards. Solution: First we find each possibility and then we multiply the results, by multiplication rule. 26 Number of ways to select 3 red cards is : = 26C3=2,600 3 Number of ways to select 3 black cards is : 26 =26C2= So together there are 2,600 x325 = 8845,000 ways of doing such a selection. Example 18: Certain class has 10 male and 8 female students. How many are there to form a committee of members consisting of: a. 5 people. b. 3 male and 2 female c. 5 people with the same gender. Solution: a. In this part we do not care about the gender of the person selected. So we could have all 5 male or female, or any combinations of the two groups. This means we want to select 5 people from a group of 18 people with no order counted. So this is clearly a combination problem and there are 18 =18C5=8,568 ways of choosing 5 5 people from 18 people. 10

11 Elementary Statistics Chapter 03 Dr. hamsary Page 11 b. Again, as we did in example17, we find each possibility and then we multiply the results, by multiplication rule. There are 10 =10C3=120 ways of choosing 3 men from 10 3 men and 8 =8C2=28 ways of choosing 2 women from 8 women. 2 So together there are 8 * 10 =28x120 = 3,360 ways of doing such a committee. 2 3 c. There are 10 =10C5=252 ways of choosing 5 men from 10 men and 5 8 8C5=56 ways 5 of choosing 5 women from 8 women. So together there are 252x 56 = 14,112 ways of doing such a committee. Example 19A: How many distinct four letter code words is possible using the letters in the word MATH. Solution: There are 4! =4*3*2*1=24 possible codes. Example 19: How many distinct four letter code words is possible using the letters in the word SHIA. Example 20A: How many distinct four letter code words is possible using the letters in the word STATISTICS? 10! Solution: There are =50400 possible codes. 3! 3! 2! 1! 1! Example 20: How many distinct four letter code words is possible using the letters in the word MISSISSIPPI? 11

12 Elementary Statistics Chapter 03 Dr. hamsary Page 12 Probability Experiment Event (Outcome) Sample Space Experiment: is a process of obtaining observations. Event: Any result of a single experiment. Sample Space: consists of all possible single events. Probability: We will denote the probability of an event E by P(E). There are two different methods to define the probability of any specific event: o Classical Rule: o Empirical Rule ( Relative Frequency): Classical Rule: Suppose we observe an experiment for n times and equally likely in which the event E occurs f times, then we have: f P( E) =. n Empirical Rule: Suppose we observe an experiment for a large number of times, not necessarily equally likely, then P(E) is defined to be the ratio of number of times E occurred to the number of times the experiment is repeated. Note that in classical approach, they require equally likely outcomes which is not always practical. 12

13 Elementary Statistics Chapter 03 Dr. hamsary Page 13 Probability Rules: For any two events E, and F we have the following rules: 1-0 P( E ) 1, If P(E) =0 then E is impossible event. If P(E) =1 then E is certain or sure event. 2-First rule of addition: P( E F ) = P( E ) + P(F ) if E and F are mutually exclusive events. E F 3-Second rule of addition: P( E F ) = P( E ) + P(F ) P( E F ) E F E F d. Complementary Events: P(E) = 1 P(E), or P( E) = 1 P( E) E E 13

14 Elementary Statistics Chapter 03 Dr. hamsary Page 14 Example 21: A single coin is flipped, then a. The experiment is to observe the up face on the coin. b. The sample space consists of Head or Tail. We usually write it as follows: S={H, T}. c. The probability of getting head is P( H) = 1 2 d. The probability of getting tail is PT ( ) = 1 2 Example 22: A single die is rolled, then a. The experiment is to observe the up face on the die. b. The sample space consists of the numbers 1 to 6. We usually write it as follows: S={1, 2, 3, 4, 5, 6}. a. The probability of observing any number is 1/6. In particular, P{ 2} 3 b. The probability of observing an odd number is P{ odds } = = 6 since the event of being odd is {1, 3, 5}. c. The probability of observing the whole sample space is P( S) = 1, since S={1, 2, 3, 4, 5, 6}. This is an example of certain event. 1 2, 1 =, 6 c. The probability of observing 7 is given by P{ 7} = 0, since it is no 7 on a single die. This is an example of impossible event. 14

15 Elementary Statistics Chapter 03 Dr. hamsary Page 15 Example 23: A single coin is flipped and A single die is rolled at the same time. a. The experiment is to observe the up face on the coin and on the die. b. The sample space will look like: S={(T,1), (T,2), (T,3), (T,4), (T,5), (T,6) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6)}. Some times it is easier to use tree diagram to write the sample space: 1 c. The probability of tail on the coin and 4 on the die is given by P{( T, 4)} =. 12 d. The probability of head on the coin and an even number on the die is given by 3 P{( H, Even)} = = 12 Example 24: Two coins are tossed together. a. The experiment is to observe the up faces 2 coins. b. The sample space will look like: 1, since {(H, even)}= {(H,2), (H, 4), (H,6)}. 4 S = { H H, HT, TH, TT } c. The probability of two heads is given by P{ HH}= 1 4. d. The probability of exactly one head is given by 2 1 P{ one H } = =, since {one H}={HT, TH}. 4 2 e. The probability of at least one head is given by P{ at least OneHead } = 3 4, since { at least one Head}={HH, HT,TH}. 15

16 Elementary Statistics Chapter 03 Dr. hamsary Page 16 Example 25: Two dice are rolled together. Find the number of all possible outcomes. a. The experiment is to observe the up faces 2 dice. b. The sample space will look like S ={ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. Since there are 6x6 = 36 possible out comes.r c. The probability of getting the sum of 7 on 2 dice is given by 6 1 P{ sum= 7} = =, since {sum = 7}={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} d. The probability of getting the sum of 11 on 2 dice is given by P{ sum= 11} = = 36 since {sum = 11}={(5,6), (6,5)}. e. The probability of getting the sum of 7 or 11 on 2 dice is given by P{ sum= 7or11} = P{ sum= 7} + P{ sum= 11} = + = = , since the events {sum=7} and {sum=11} are mutually exclusive.. f. P{ sum = 8or double} = P{ sum = 8} + P{ double} P { sum = 8} { bouble} g, since they overlap, then by addition rule we have the above formula. Also we have {sum=8}={(2,6), (3,5), (4,4), (5,3), (6,2)}, {double}={(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}, and {sum =8} {doble}={(4,4)}. Hence we have: P{ sum = 8or double } = + = = b 1 18, 16

17 Elementary Statistics Chapter 03 Dr. hamsary Page 17 Example 26: In a family of 4 children let stands for boy and for girl. Then sample space will have 2*2*2*2= 2 4 = 16 possibilities and it look like the following table: First Second Third Fourth a. The probability of all 4 children are boy is given by: P{ All oys}= 1, since only one of the 16 possible outcomes is all boy, b. The probability of exactly one boy is given by P{ Exactly one boy}= = 16 since there are 6 single events that consists of exactly one boy,,,,. c. The probability of at least one by is given by: 1 15 P{ At least one boy} = 1 P{ No boy} = 1 =, indeed there are 15 cases out of possible outcomes that are at least one boy. d. The probability of same sex ( all are boys or all are girls) is given by P{ Same sex} = P{ all boys or all girls} = + = =, since the two events of E={all boys}, and F={all girls} are mutually exclusive. 1 4, 17

18 Elementary Statistics Chapter 03 Dr. hamsary Page 18 Example 27: A card is drawn at random from a deck of 52 cards. Find the probability of the following events: a. The event of selecting a king. b. The event of selecting a heat. c. The event of selecting a king of heart. d. The event of selecting a king or a heart. Solution: let us define: K = the event of the selected card is a King. H = the event of the selected card is a Heart. Example 28: In the following table we have two factors gender and smoking, as shown. Smoking No Smoking Total Male Female Total If one person is selected at random, then: a. Find the probability that the selected person is a man b. Find the probability that the selected person is a smoker c. Find the probability that the selected person is a man smoker d. Find the probability that the selected person is a man or smoker. Solution: 18

19 Elementary Statistics Chapter 03 Dr. hamsary Page 19 Example 29: An unfair die is rolled once and the upward face is recorded. The faces have the following probabilities: Face Probability a. What is the probability of rolling an even number? b. What is the probability of rolling something other than a two? c. If this were a fair die, what would be the probability of rolling each face? Solution: Example 30: If P(A) = 0.25 and P() = a. Find the probability P(A or ) if P(A and ) = 0.15 b. P(A and ) if P(A or ) = 0.60 c. P(A or ) if A and are mutually exclusive Solution: P(A ) = P(A) + P() P(A ) 19

20 Elementary Statistics Chapter 03 Dr. hamsary Page 20 Multiplication of probabilities Independent Events Dependent Events Conditional Probability (ayes s Rule) Independent Events: Two events A and are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other one. Otherwise A and are dependent. The probability of A and, which is denoted by P(A and ) or P(A ), is defined by o P(A ) = P(A)P(), if A and are independent o P(A ) = P(A)P( A), if A and are dependent. P( A)is read as conditional probability of A given. Conditional Probability: From the above we can write P( A ) = P( A) P( A) and from this we have: P(A ) P( A) =, which is also known as ayes s rule. P(A) Question: How to check two events are independent? = or P( A) = P( ) Answer: If P( A ) P( A)P( ) ecause if A and are independent, then P(A ) = P(A)P(). Substitute this in the above, ayes s rule formula and then we have ( ) ( ) ( ) P( A ) P A P P( A) = = = P( ), and similarly P( A) P A ( ) P(A ) = P A 20

21 Elementary Statistics Chapter 03 Dr. hamsary Page 21 Example 31: A single coin is flipped and A single die is rolled at the same time. The sample space will look like: S={(T,1), (T,2), (T,3), (T,4), (T,5), (T,6) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6)}. Then we have: PT {, 3) = 1 1 PTP ( ) ( 3) = = 12, which is an indication of the events { T } and the { 3 } are independent. Also we have 1 P{ T, 3} 1 P{T 3 ) = == 12 = = P{ T}, this is exactly the probability of getting tail on P{} a single coin. Thus P{T 3 ) P{ T} coin and 3 on the die. Example 32: Two coins are tossed together. 1 1 Then we have P{ HH} = P{ H} P{ H} = = 2 2 are independent. =,which is another way of showing independence of tail on the 1. This equation means the outcomes of the 2 coins 4 Example 33: A box contains 5 red balls and 7 blue balls. We select 2 balls at random and find the probability of choosing two red balls under the following situations: Sampling With Replacement: We replace the first ball before we take the second one. Sampling Without Replacement: We do not replace the first ball before we take the second one. Solution: Let R 1 = The event of selecting the first ball is Red. Let R 2 = The event of selecting the second ball is Red. 21

22 Elementary Statistics Chapter 03 Dr. hamsary Page 22 In the first case R 1 and R 2 are independent, sine the occurrence of R 1 will not change the probability of the occurrence of R 2. Hence we have: 5 5 P( R1R 2 ) = P( R 1)P( R 2 ) = = In the second case R 1 and R 2 are dependent, sine the occurrence of R 1 will change the probability of the occurrence of R 2. Hence we have: P( R1R 2 ) = P( R 1)P( R 2 R 1) = = , we know that P( R1 ) =, and P( R2 ) = since we will not replace the first ball back in to the box. 11 Example 34: Repeat example 3 with 3 balls. Solution: Let R 1 = The event of selecting the first ball is Red. Let R 2 = The event of selecting the second ball is Red. Let R 3 = The event of selecting the third ball is Red. In the first case R 1, R2 and R 3 are all independent, sine the occurrence of R 1 will not change the probability of the occurrence of R 2 and so on. Hence we have: P( R1R2R 3 ) = P( R 1)P( R 2 )P( R 3 ) = = In the second case R 1, R2 and R 3 are dependent, sine the occurrence of R 1 will change the probability of the occurrence of R 2 and R 3. Hence we have: P( R1R2R 3 ) = P( R 1)P( R 2 R 1)P( R 3 R1R 2 ) = =

23 Elementary Statistics Chapter 03 Dr. hamsary Page 23 Example 35: In the following table we have two factors gender and smoking, as shown. Smoking No Smoking Total Male Female Total If one person is selected at random, then a. Probability of choosing a man smoker. b. iven we have selected a man, what is the probability that he is smoker? c. iven we have selected a smoker, what is the probability that the person is man? d. iven we have selected a smoker, what is the probability that the person is a women? Solution: Let M = the event of selecting a Man, S= the event of selecting a Smoker, and finally W = the event of selecting a Women. a. 240 P( MS ) = P( M S ) = = b. Note that this is conditional probability, since we are given some information about the event M. So we have, by the conditional probability formula: P( S M ) 240 / 500 P(S M) = = = 080., or divide the number of male smokers by total P( M ) 300 / 500 number of male (240/300 =0.60). 240 P(S M) = = c. This part is also conditional probability as follows: P( M S ) 240 / 500 P(M S) = = or divide the number of male smokers by total P( S ) 360 / 500 number of smokers (240/360 =0.667). 23

24 Elementary Statistics Chapter 03 Dr. hamsary Page P(M S) = d. Again we have conditional probability of selecting a women out of only smokers. P(W S ) 120 / 500 P(W S) = = P( S ) 360 / P(W S ) = Example 36: You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. a. Are the outcomes on the two cards independent? Why? b. Find P( a heart on the 1 st card club on the 2 nd ). c. Find P(a club on the 1 st card and a heart on the 2 nd ). d. Find the probability of drawing a club and a 3 heart in either order. Solution: a. No, they are not independent. The probability of the second card depend on the first card drawn, since the first card was not replaced in the deck. b. P (a heart on the 1 st card club on the 2 nd )= c. P(a club on the 1 st card and a heart on the 2 nd )= d. P(probability of drawing a club and a heart in either order)= P (a heart 1 st and club on the 2 nd )+ P(a club 1 st card and a heart on the 2 nd )=

25 Elementary Statistics Chapter 03 Dr. hamsary Page 25 Example 37: Suppose in a large city 10% of people have certain virus. We select 5 persons at random and find the probability that: a. All of them have the virus. b. None of them have the virus. c. At least one of them has the virus. Solution: Let V 1 = the event that the first person has the virus, V 2 = the event that the second person has the virus, and so on. a. P ( all 5 have the virus) = PVVVVV ( ) = PV ( ) PV ( )... PV ( ) = ( )( )( )( )( ) ( ) = = b. If P (Virus) =01.10, P( No Virus) = =0.90, P (None of them has the Virus) = (0.90)(0.90)(0.90)(0.90)(0.90) = c. We use the complementary events rule: P (not E) = 1-P (E), so we have: P( At least one has the Virus) = 1-P(No one has the Virus) = = Example 38: A bag contains 7 red and 13 blue m&m candies and we randomly select 2 cadies without replacement. a. What is the probability of choosing both red candies? b. What is the probability of choosing at least one red? Solution: a. Let us do it by combination formula. As we know 20 20! 20C 2 = = =190 possible 2 2! ( 20 2)! ways of selecting 2 candies from total of 20 candies. 25

26 Elementary Statistics Chapter 03 Dr. hamsary Page 26 Also there are 7 7! 7C2= = =21 possibility of selecting 2 red from 7 red candies. So 2 2! ( 7 2)! the probability is given by: P ( 2red ) = = b. To find the probability of at least one red we do the complimentary probability as follows: P( at 13 13! least one red )= 1- P( no red)=1- P(both blue) There are 13C 2 = = =78 ways 2 2! ( 13 2)! of selecting 2 blue candies from a set of 13 blue candies. P ( 2blue) = = = P( at least one red )= 1- P( no red)=1- P(both blue)= =0.598 Example 39: A bag contains 7 red and 13 blue m&m candies and we randomly select 5 cadies without replacement. a. What is the probability of choosing both 2 red and 3 blue candies? b. What is the probability of choosing at least one red? Solution: a. Let us do it by combination formula. As we know 20 20! 20C5 = = = ! ( 20 5)! possible ways of selecting 5 candies from total of 20 candies. 26

27 Elementary Statistics Chapter 03 Dr. hamsary Page 27 Also there are 7 7! 7C2 = 2 = =21 possibility of selecting 2 red from 7 red candies. 2! ( 7 2)! 13 13! Similarly13C 3 = = = ! ( 13 3)! So the probability is given by: P ( 2red, 3blue) * 286 = = a. Again to find the probability of at least one red we do the complimentary probability as follows: P( at least one red )= 1- P( no red)=1- P(both blue) There are 13 13! 13C 5 = = =1287ways of selecting 2 blue candies. 5 5! ( 13 5)! P ( 5blue) = = = P (at least one red )= 1- P( no red)=1- P(both blue)= =

28 Elementary Statistics Chapter 03 Dr. hamsary Page 28 Review A standard deck of cards contains 52 cards. There are 4 suits in the deck: 13 hearts, 13 diamonds, 13 clubs, 13 spades. Each suit has one ace, one two, one three,, one jack, one queen, and one king. Consider the probability experiment of drawing one card from a deck of cards. a. Determine the sample space of the probability experiment. b. Determine the probability of randomly selecting the king of hearts from the deck of cards. Interpret this probability. c. Determine the probability of randomly selecting any heart from a deck of cards. Interpret this probability. d. Determine the probability of randomly selecting a king from a deck of cards. Interpret this probability. 2. Exclude leap years from the following calculations. Interpret each probability. a. Determine the probability that a randomly selected person ahs a birthday on the 1 st day of a month. b. Determine the probability that a randomly selected person has a birthday on the 31st day of a month. c. Determine the probability that a randomly selected person was born in December. d. Determine the probability that a randomly selected person has a birthday on November 8. e. If you just met somebody and she asked you to guess her birthday, are you likely to be correct? f. Do you think it is appropriate to use the methods of classical probability to compute the probability that a person is born in December? 3. In a national survey conducted by the Center for Disease Control in order to determine college students healthrisk behaviors, college students were asked: How often do you wear a seat belt when riding in a car driven by someone else? The frequencies appear in the following table: Response Never Rarely Sometimes Most of the Time Always Frequency ,257 2,518 a. Approximate the probability that a randomly selected college student never wears a seat belt when riding in a car driven by someone else. 28

29 Elementary Statistics Chapter 03 Dr. hamsary Page 29 b. Would you consider it unusual to find somebody who never wears a seat belt when riding in a car driven by someone else? Why? c. Approximate the probability that a randomly selected college student sometimes wears a seat belt when riding in a car driven by someone else. Interpret this probability. 4. In Problem a-e, find the probability of the indicated event if P(A) = 0.20 and P() = a. P(A or ) if P(A and ) = 0.30 b. P(A and ) if P(A or ) = 0.40 c. P(A or ) if A and are mutually exclusive d. P(A and ) if A and and mutually exclusive e. P ( A) f. P ( ) 5. A standard deck of cards contains 52 cards. One card is randomly selected from the deck. a. Compute the probability of randomly selecting a heart or club from a deck of cards. b. Compute the probability of randomly selecting a heart or club or diamond from the deck of cards. c. Compute the probability of randomly selecting an ace or heart from a deck of cards. 6. You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. a. Are the outcomes on the two cards independent? Why? b. Find P(3 on the 1 st card and 10 on the 2 nd ). c. Find P(10 on the 1 st card and 3 on the 2 nd ). d. Find the probability of drawing a 10 and a 3 in either order. 7. Diagnostic tests of medical conditions have several results. The test result can be positive or negative, whether or not a patient has the condition (+ indicates a patient has the condition). Consider a random sample of 200 patients, some of whom have a medical condition and some of whom do not. Results of a new diagnostic test for the condition are shown. Condition Present Condition Absent Row Total Test Result Test Result Column Total Assume the sample is representative of the entire population. For a person selected at random, compute the following probabilities: 29

30 Elementary Statistics Chapter 03 Dr. hamsary Page 30 a. P(+, given condition present); this is known as the sensitivity of a test. b. P(-, given condition present); this is known as the false-negative rate. c. P(-, given condition absent); this is known as the specificity of a test. d. P(+, given condition absent); this is known as the false-positive rate. e. P(condition present and +); this is the predictive value of the test. f. P(condition present and -). 8. To emphasize the importance of correct shelving of books, a librarian tells a group of students the number of possible orders in which just 6 books may be placed on a shelf. What is that number? 9. During a Compute Daze special promotion, a customer purchasing a compute and printer is given a choice of three free software packages. There are 10 different software packages from which to select. How many different groups of software packages can be selected? 10. One professor grades homework by randomly choosing 5 out of 12 homework problems to grade. a. How many different groups of 5 problems are there from the 12 problems? b. Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded? c. Silvia did 7 problems. How many different groups of 5 did she complete? What is the probability that one of the groups of 5 she completed comprised the group selected to be graded? 30