Probability Section 19 NOTE: We are skipping Section 18. 2/21/2019 MATH 106, Section 19

Probability Notation The subject of probability is the study of how we can predict the likelihood of the occurrence of some event. The probability of an event occurring is the percentage (proportion) of times that the event will occur in a large number of trials. Illustration: Roll a standard, fair, six-sided die, and observe the number of spots facing upward. The probability of observing a multiple of 3 is , and a correct interpretation of this probability is 1/3 that every time the die is rolled three times, exactly one of the outcomes will be a multiple 3. that as the die is rolled many times, the percentage (proportion) of outcomes which will be a multiple 3 tends to become closer to 33.333…% (0.333…) . 2/21/2019 MATH 106, Section 19

This is not correct, since 2 P(E) = — 11 If U is a set of all possible outcomes, then U is called the sample space, and any subset E is called an event. If U is a (finite) sample space consisting of outcomes which are equally likely, and E is any subset of U, then the probability of event E is defined to be #E P(E) = —– #U Illustrations: Roll a standard, fair, six-sided die, and observe the number of spots facing upward. U = E = “less than 4” = Roll two standard, fair, six-sided dice, and observe the number of spots facing upward. 3 1 P(E) = — = — 6 2 {1, 2, 3, 4, 5, 6} {1, 2, 3} This is not correct, since 2 P(E) = — 11 {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} {2, 3} the outcomes in U are not equally likely! 2/21/2019 MATH 106, Section 19

{all pairings of two cards, repetition possible} #U = #E = 522 #1 A standard deck of 52 cards contains 13 cards of each of the four suits (club, diamond, heart, spade), and each suit contains 13 denominations (two, three, four, five, six, seven, eight, nine, ten, jack, queen, king, ace). (a) (b) One card is randomly selected from the deck. What is the probability that the card is a heart? U = E = {all cards} #U = #E = 52 #E P(E) = —– = #U 13 1 — = — = 25% 52 4 {hearts} 13 Two cards are randomly selected from the deck by first selecting one card, replacing that card in the deck, and selecting the second card. What is the probability that both cards are hearts? U = E = {all pairings of two cards, repetition possible} #U = #E = 522 {all pairings of two hearts, repetition possible} 132 2/21/2019 MATH 106, Section 19

{all pairings of two cards, repetition possible} #U = #E = 522 Two cards are randomly selected from the deck by first selecting one card, replacing that card in the deck, and selecting the second card. What is the probability that both cards are hearts? U = E = {all pairings of two cards, repetition possible} #U = #E = 522 {all pairings of two hearts, repetition possible} 132 #E P(E) = —– = #U 169 —— = 0.0625 or 6.25% 2704 2/21/2019 MATH 106, Section 19

{all pairs (combinations) of two cards} #U = #E = C(52,2) Two cards are randomly selected from the deck by first selecting one card, and then selecting the second card from the 51 cards that remain in the deck. What is the probability that both cards are hearts? U = E = {all pairs (combinations) of two cards} #U = #E = C(52,2) {all pairs (combinations) of two hearts} C(13,2) #E P(E) = —– = #U 78 —— = 0.0588 or 5.88% 1326 A five-card poker hand is randomly selected from the deck. What is the probability that the hand contains exactly 2 aces? U = E = {all hands (combinations) of five cards} {all hands (combinations) of 2 aces and 3 non-aces} 2/21/2019 MATH 106, Section 19

{all hands (combinations) of five cards} A five-card poker hand is randomly selected from the deck. What is the probability that the hand contains exactly 2 aces? U = E = {all hands (combinations) of five cards} {all hands (combinations) of 2 aces and 3 non-aces} #U = #E = C(52,5) C(4,2)  C(48,3) #E P(E) = —– = #U 103,776 ————— = 0.0399 or 3.99% 2,598,960 2/21/2019 MATH 106, Section 19

{all hands (combinations) of five cards} #U = #E = C(52,5) A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of all spades (which is a spade flush)? U = E = {all hands (combinations) of five cards} #U = #E = C(52,5) {all hands (combinations) of five spades} C(13,5) #E P(E) = —– = #U 1287 ————— = 0.0005 or 0.05% 2,598,960 A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of all one suit (which is a flush)? C(52,5) U = E = {all hands (combinations) of five cards} #U = #E = {all hands (combinations) of five clubs or five diamonds or five hearts or five spades} 2/21/2019 MATH 106, Section 19

{all hands (combinations) of five cards} A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of all one suit (which is a flush)? U = E = {all hands (combinations) of five cards} {all hands (combinations) of five clubs or five diamonds or five hearts or five spades} #U = #E = C(52,5) C(13,5) + C(13,5) + C(13,5) + C(13,5) = 4C(13,5) #E P(E) = —– = #U 5148 ————— = 0.0020 or 0.20% 2,598,960 2/21/2019 MATH 106, Section 19

{all hands (combinations) of five cards} (g) (h) A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of three cards of the same denomination (three of a kind) together with two cards of second denomination (two of a kind)? (Such a hand is called a full house.) U = E = {all hands (combinations) of five cards} {all hands (combinations) of three of a kind and two of a kind} #E P(E) = —– = #U 3744 ————— = 0.0014 or 0.14% 2,598,960 #U = #E = C(52,5) C(13,1)C(4,3)C(12,1)C(4,2) A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of two cards from one suit, two cards from a second suit, and the remaining card from a third suit? U = E = {all hands (combinations) of five cards} 2/21/2019 MATH 106, Section 19

{all hands (combinations) of five cards} A five-card poker hand is randomly selected from the deck. What is the probability that the hand consists of two cards from one suit, two cards from a second suit, and the remaining card from a third suit? U = E = {all hands (combinations) of five cards} {all hands (combinations) of two cards from one suit, two cards from a second suit, and the remaining card from a third suit} #U = #E = C(52,5) C(4,1)C(13,2)C(3,1)C(13,2)C(2,1)C(13,1) #E P(E) = —– = #U 1,898,208 ————— = 0.730 or 73.0% 2,598,960 2/21/2019 MATH 106, Section 19

In Section 19 Homework Problems #2, #3, #4, and #5, Homework Hints: In Section 19 Homework Problems #2, #3, #4, and #5, the approach that can be used is at least one of the approaches used in the different parts of #1 on the Section #19 Handout. Quiz #5 COMING UP! Be sure to do the review problems for this, quiz posted on the internet. The link can be found in the course schedule. 2/21/2019 MATH 106, Section 19

A fair coin is flipped eight times. (a) #2 A fair coin is flipped eight times. (a) (b) What is the probability of observing exactly six heads? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of 6 Hs and 2 Ts} #U = #E = 28 = 256 #E P(E) = —– = #U 28 —— = 0.109 or 10.9% 256 C(8,6) = 28 What is the probability of observing at least six heads? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with at least 6 Hs} #U = #E = 28 = 256 #E P(E) = —– = #U 37 —— = 0.145 or 14.5% 256 C(8,6) + C(8,7) + C(8,8) = 37 2/21/2019 MATH 106, Section 19

What is the probability of observing at most six heads? (c) (d) What is the probability of observing at most six heads? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with at most 6 Hs} #U = #E = 28 = 256 #E P(E) = —– = #U 247 —— = 0.965 or 96.5% 256 28 – [C(8,7) + C(8,8)] = 247 What is the probability of observing exactly two heads? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of 2 Hs and 6 Ts} #U = #E = 28 = 256 #E P(E) = —– = #U 28 —— = 0.109 or 10.9% 256 C(8,2) = 28 2/21/2019 MATH 106, Section 19

What is the probability of observing at least three heads? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with at least 3 Hs} #U = #E = 28 = 256 #E P(E) = —– = #U 28 – [C(8,0) + C(8,1) + C(8,2)] = 219 219 —— = 0.855 or 85.5% 256 What is the probability of observing no consecutive heads, that is, no heads are adjacent in the eight flips? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with no adjacent Hs} #U = #E = 28 = 256 #E P(E) = —– = #U 55 —— = 0.215 or 21.5% 256 F(8) = 55 2/21/2019 MATH 106, Section 19

{all strings of size 8 consisting of Hs and Ts} What is the probability of observing no consecutive tails, that is, no tails are adjacent in the eight flips? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with no adjacent Ts} #U = #E = 28 = 256 #E P(E) = —– = #U 55 —— = 0.215 or 21.5% 256 F(8) = 55 What is the probability of observing no consecutive heads and no consecutive tails, that is, no heads are adjacent in the eight flips and no tails are adjacent in the eight flips? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with no adjacent Hs and no adjacent Ts} 2/21/2019 MATH 106, Section 19

{all strings of size 8 consisting of Hs and Ts} What is the probability of observing no consecutive heads and no consecutive tails, that is, no heads are adjacent in the eight flips and no tails are adjacent in the eight flips? U = E = {all strings of size 8 consisting of Hs and Ts} {all strings of size 8 consisting of Hs and Ts with no adjacent Hs and no adjacent Ts} = {HTHTHTHT , THTHTHTH} #U = #E = 28 = 256 #E P(E) = —– = #U 2 —— = 0.008 or 0.8% 256 2 2/21/2019 MATH 106, Section 19

{all permutations of the first-graders} #3 Nine first-graders, consisting of five girls and four boys, enter a park. (a) (b) All of the first graders are randomly assigned to sit in a row on a bench. What is the probability that Cindy and Dani will sit next to each other? U = E = {all permutations of the first-graders} {all permutations of the first-graders with Cindy & Dani together} #U = #E = P(9,9) = 9! = 362,880 #E P(E) = —– = #U 80,640 ———– = 0.222 or 22.2% 362,880 P(8,8)P(2,2) = 80,640 All of the first graders are randomly assigned to sit in a row on a bench. What is the probability that all the boys will sit together? U = E = {all permutations of the first-graders} {all permutations of the first-graders with all the boys together} 2/21/2019 MATH 106, Section 19

{all permutations of the first-graders} (b) All of the first graders are randomly assigned to sit in a row on a bench. What is the probability that all the boys will sit together? U = E = {all permutations of the first-graders} {all permutations of the first-graders with all the boys together} #U = #E = P(9,9) = 9! = 362,880 #E P(E) = —– = #U 17,280 ———– = 0.048 or 4.8% 362,880 P(6,6)P(4,4) = 17,280 2/21/2019 MATH 106, Section 19

{all permutations of the first-graders with all the boys together} (c) (d) All of the first graders are randomly assigned to sit in a row on a bench with all the boys sitting together. What is the probability that Cindy and Dani will sit next to each other? U = E = {all permutations of the first-graders with all the boys together} {all permutations of the first-graders with all the boys together and with Cindy & Dani together} P(6,6)P(4,4) = 17,280 #E P(E) = —– = #U 5,760 ———– = 0.333 or 33.3% 17,280 #U = #E = P(5,5)P(4,4)P(2,2) = 5760 All of the first graders are randomly assigned to sit in a row on a bench with all the boys sitting together. What is the probability that all the girls will sit together? U = E = {all permutations of the first-graders with all the boys together} {all permutations of the first-graders with all the boys together and with all the girls together} 2/21/2019 MATH 106, Section 19

{all permutations of the first-graders with all the boys together} All of the first graders are randomly assigned to sit in a row on a bench with all the boys sitting together. What is the probability that all the girls will sit together? U = E = {all permutations of the first-graders with all the boys together} {all permutations of the first-graders with all the boys together and with all the girls together} #U = #E = P(6,6)P(4,4) = 17,280 P(2,2)P(5,5)P(4,4) = 5760 #E P(E) = —– = #U 5,760 ———– = 0.333 or 33.3% 17,280 2/21/2019 MATH 106, Section 19

{all permutations of the first-graders} All of the first graders are randomly assigned to sit in a row on a bench. What is the probability that all the boys will sit together and all the girls will sit together? U = E = {all permutations of the first-graders} {all permutations of the first-graders with all the boys together and all the girls together} #E P(E) = —– = #U 5760 ———– = 0.016 or 1.6% 362,880 #U = #E = P(9,9) = 9! = 362,880 P(2,2)P(5,5)P(4,4) = 5760 All of the first graders are randomly assigned to sit in a circle on a merry-go-round. What is the probability that all the girls will sit together? U = E = {all circle arrangements of the first-graders} {all circle arrangements of the first-graders with all the girls together} 2/21/2019 MATH 106, Section 19

{all circle arrangements of the first-graders} All of the first graders are randomly assigned to sit in a circle on a merry-go-round. What is the probability that all the girls will sit together? U = E = {all circle arrangements of the first-graders} {all circle arrangements of the first-graders with all the girls together} #U = #E = (9 – 1)! = 40,320 #E P(E) = —– = #U 2880 ——— = 0.071 or 7.1% 40,320 (5 – 1)!  5! = 2880 2/21/2019 MATH 106, Section 19

{all circle arrangements of the first-graders} All of the first graders are randomly assigned to sit in a circle on a merry-go-round. What is the probability that all the boys will sit together? U = E = {all circle arrangements of the first-graders} {all circle arrangements of the first-graders with all the boys together} #U = #E = (9 – 1)! = 40,320 #E P(E) = —– = #U 2880 ——— = 0.071 or 7.1% 40,320 (6 – 1)!  4! = 2880 All of the first graders are randomly assigned to sit in a circle on a merry-go-round. What is the probability that Cindy and Dani will sit next to each other? U = E = {all circle arrangements of the first-graders} {all circle arrangements of the first-graders with Cindy and Dani together} 2/21/2019 MATH 106, Section 19

{all circle arrangements of the first-graders} All of the first graders are randomly assigned to sit in a circle on a merry-go-round. What is the probability that all the girls will sit together? U = E = {all circle arrangements of the first-graders} {all circle arrangements of the first-graders with Cindy and Dani together} #U = #E = (9 – 1)! = 40,320 #E P(E) = —– = #U 10,080 ——— = 0.250 or 25.0% 40,320 (8 – 1)!  2! = 10,080 2/21/2019 MATH 106, Section 19

{all permutations of the 365 dates taken 30 at a time} #4 Several people are in a room, and we are interested in the probability that two or more people have the same birthday. We shall obtain this probability by focusing on the probability of the opposite event, that is, the probability that every person in the room has a different birthday. We shall ignore Feb 29, and treat the other 365 dates as equally likely birthdays. In other words, we can assume that each person in the room has been randomly assigned one of the 365 dates with repetition possible. (a) If there are 30 people in the room, what is the probability that two or more people have the same birthday? U = E = ~E = {all assignments of the 365 dates to the 30 people, repetition possible} {all assignments of the 365 dates to the 30 people, with at least one repetition} {all permutations of the 365 dates taken 30 at a time} 2/21/2019 MATH 106, Section 19

{all permutations of the 365 dates taken 30 at a time} If there are 30 people in the room, what is the probability that two or more people have the same birthday? U = E = ~E = {all assignments of the 365 dates to the 30 people, repetition possible} {all assignments of the 365 dates to the 30 people, with at least one repetition} {all permutations of the 365 dates taken 30 at a time} #U = #E = #~E = 36530 36530 – P(365,30) P(365,30) #E P(E) = —– = #U 36530 – P(365,30) ———————– = 0.706 or 70.6% 36530 2/21/2019 MATH 106, Section 19

{all permutations of the 365 dates taken n at a time} (b) Suppose there are n people in the room, and the probability that two or more people have the same birthday is larger than 1/2. What is the smallest n could be? U = E = ~E = {all assignments of the 365 dates to the n people, repetition possible} {all assignments of the 365 dates to the n people, with at least one repetition} {all permutations of the 365 dates taken n at a time} #U = #E = #~E = 365n 365n – P(365, n) P(365, n) 0.476 or 47.6% if n = 22 #E P(E) = —– = #U 365n – P(365, n) ———————– = 365n 0.507 or 50.7% if n = 23 0.538 or 53.8% if n = 24 2/21/2019 MATH 106, Section 19

In Section 19 Homework Problems #6, #7, #8, and #9, Homework Hints: In Section 19 Homework Problems #6, #7, #8, and #9, the approach that can be used is at least one of the approaches used in the different parts of #2 and #3 on the Section #19 Handout. 2/21/2019 MATH 106, Section 19