Probability Hw Solutions (5)

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Probability Hw Solutions (5)

  1. 1. Ms. Jaffe Precalculus Homework due Friday 9/5 (11B, 304), Monday 9/8 (11A) NAME: _____________________________________________________ 1. Review The Lottery by Shirley Jackson. What is the difference between the probability existent within this story and the probability used in playing Deal or No Deal? The probability in “The Lottery” can be viewed as either dependent or independent. Regardless, the probability of choosing a black dot does not change for each person at any given time. In Deal or No Deal the events are dependent. You are also considering more than one probability, the probability of your choosing a briefcase with a high value and the probability your original briefcase holds a high value. 2. Describe the difference between two events that are dependent and two events that are independent. Give an example of each. Two events are dependent if the outcome of the second eventdepends on the outcome of the first. An example would be removing two cards from a standard deck of 52 cards without replacement. After I choose the first card, the number of possibilities there are for the second card decreases by one. Two events are independent if the outcome of the second event does not change based on the outcome of the first. An example would be tossing two coins. The fact that the first lands on heads has no effect on whether or not the second will land on heads. 3. Does The Lottery describe dependent or independent events? Explain. We can make an argument for either. We can say independent because no one looks at their paper until everyone has chosen. We can say dependent because the last person only has one piece of paper to choose from. Please keep in mind that this is a complex problem and is purely for discussion purposes. 4. What other games involve probability? Describe the game and explain where probability is involved? All card games, most game show games, most board games, etc. 5. What is the probability of getting 750,000 dollars in a briefcase if we can choose two briefcases out of seven? (I only asked one class this) The probability is 2/8. There are actually 8 briefcases because we have to include the one you chose at the very beginning. The probability for each individual briefcase having 750,000 is therefore 1/8. When we choose two it is the probability of the first or the second which is 1/8 + 1/8 = 2/8.
  2. 2. Ms. Jaffe Precalculus Homework due Tuesday 9/9 (all classes) NAME: _________________________________________________________ 1. Give an example of a problem which would require the use of permutations and solve. What is the number of possible locker combinations for a Masterlock. (Notice the misuse of the word combinations. In a lock the order of the numbers matters.) 40P3 = 40 = 64000 3 Note that we can repeat the same number,so this is a permutation with replacement. The formula is nPr = nr. 2. Give an example of a problem which would require the use of combinations and solve. How many different groups of 4 can I create in a class of 28. 28C4 = 20475 3. How does the formula for permutations and combinations change if there is replacement? The formula for permutations becomes nr because for each slot there exists the same number of possibilities. The formula for combinations will result in a similar change. It is far more complex so we will leave this for a later discussion. 4. How many seven-digit phone numbers are there if the first digit cannot be a zero, the last digit is even, and the third digit is less than 5? The numbers below represent the number of possibilities for each digit. Please note that 0 is even. 9*10*5*10*10*10*5 = 2,250,000 5. How many license plates are there if the license plate has three letters and three numbers? There are 26 possibilities for each of the first three spots because there are 26 letters in the alphabet and we can use a letter more than once. There are 10 possibilities for each of the three remaining spots because there are 10 digits (0-9) and we can use a digit more than once. You will not use 36 possibilities because the first spot can only have a letter or a number. If we put 36 for all spots it would mean that it is possible to have more than 3 letters / numbers. 26*26*26*10*10*10 = 17,576,000 6. How is the formula for calculating the number of permutations different from that for calculating the number of combinations? Why does this difference in the formula work? The formula for calculating combinations is the same as the formula for calculating permutations except we divide by r!. r! tells us the number of repetitions for each permutation. We divide so we rid ourselves of the repetitions of every permutation.
  3. 3. Ms. Jaffe Precalculus Homework due Wednesday 9/10 (all classes) NAME: _______________________________________________ 1. Find the number of permutations in the word MISSISSIPPI. Explain the formula for acquiring this number. We have two options for calculating this: 11! 1. 4!4!2! = 34650 The 11! Represents the total number of ways we can arrange 11 letters. The 4! represents the number of ways we can arrange the 4 Ss for each permutation; the second 4! represents the number of arrangements of the 4 Is; the 2! represents the number of arrangements of the 2 Ps. The 4!4!2! represents the number of repetitive permutations for each permutation. 2. 11C4 * 7C4 * 3C1 * 2C2 = 34650. The 11C4 represents the choice of 4 spots for the 4 Is. We are using combinations because once we choose the 4 spots, the order in which we put in the Is does not matter. This goes for the other three combinations as well. The 7C4 represents the spots chosen for the 4 Ss. We only had 7 spots remaining because 4 were already used for the Is and we have 4 Ss so we are choosing 4 spots. The 3C1 represents the number of spots available for the 1 M. The 2C2 represents the number of ways we can choose the 2 remaining spots for the 2 Ps. Please note we did not have to place the four letters into the arrangement int his order. Our combinations would change if we changed the order, but the final answer would remain the same. 2. a) There are 28 students in a class. The class wants to appoint three people to be official “leaders” of the class. How many groups of leaders are possible? Order does not matter because there is no way to distinguish between the three leaders. 28C3 = 3276 b) If the class specifically wanted a president, vice-president, and secretary, would the number of possibilities change? How and why? Order matters because we are given each person a different task. 28P3 = 19656 3. Four cards are drawn from a pack of 52 cards. Determine the probability that a) all are aces In the denominator we have the total number of ways to choose four cards from a deck = 52C4. In the numerator we have the number of ways we can choose four aces from a deck of 52 cards. 4 C4 1 There are four aces in the deck and we are choosing all four of them = 4C4. 52 C 4 = 270725 b) they are A, K, Q, J Our denominator remains the same. In our numerator we have the number of ways to choose an A, K, Q, J. We have four aces and we are choosing 1 of them, 4 kings and we are choosing 1, 4 queens and we are choosing 1, and 4 jacks and we are choosing 1 = 4C1 * 4C1*4C1*4C1.
  4. 4. C1 *4 C1 *4 C1 *4 C1 256 4  52 C 4 270725 c) they are of the same suit Let’s first do find the probability of choosing all diamonds. The denominator is still the same. In the numerator we have 13 diamonds to choose form and we are choosing 4 = 13C4; so the 13 C 4 probability is 52 C 4 . Since the number of cards in each sui is the same the probability of t choosing 4 hearts is the same as the probability of choosing 4 clubs which is the same as the probability of choosing 4 spades which is the same as the probability of choosing 4 diamonds. We 13 C 4 13 C 4 13 C 4 13 C 4 can choose 4 diamonds or 4 hearts or 4 clubs or 4 spades = 52 C 4 + 52 C 4 + 52 C 4 + 52 C 4 = 4* C4 2860 13  52 C 4 270725 4. When we calculate the number of combinations why does dividing by r! remove all the repetitive permutations? r! represents the number of repetitions of each permutation where we are choosing r objects because there are r! ways to arrange r objects. We need to get rid of the repetitions of all permutations so we divide. 5. How do the events determine the kind of tree diagram we use? Explain. An event in which the probability of each outcome is the same will use aregular tree diagram. An event in which the probability of each outcome is different will use a probability tree diagram. Please note that a regular tree diagram is a probability tree diagram, but it is unnecessary to put in the probabilities because we can determine the probability from the tree. All regular tree diagrams have independent events, but all independent events do not use regu tree diagrams. lar Dependent events can yield either tree diagram depending on the probabilities of each outcome as described above. 6. Explain when and why the Fundamental Counting Principle holds true. The Fundamental Counting Principle says if event A occurs in m ways and event B occurs in n ways then the total number of ways event A and B can occur is in m*n ways. Consider choosing an outfit given 5 shorts and 3 pairs of pants. For each short there are 3 possible pairs of pants. This occurs 5 times so we have 5 groups of 3, which is 5*3. This is the very definition of multiplication. m groups of n objects yields m*n objects. 7. I forgot my best friend’s phone number. The only thing I remember is that the second digit is a 4, the third digit is either 7 or 9, and the last digit is even. What is the probability that I dial the correct seven-digit number? The numbers below represent the number of possibilities of for each slot. 10*1*2*10*10*10*5 = 100,000. This is the total number of numbers you can dial. You only have one desired number because otherwise you will not get in touch with your best friend. Therefore, the probability is 1/100,000
  5. 5. 8. Place the letters ABCD in a circle. Write all of the different circular arrangements for these letters. There are 6 possibilities that will yield different permutations. I will leave this one to you J
  6. 6. Ms. Jaffe Precalculus Homework due Thursday 9/11 (all classes) NAME: _____________________________________________________ 1. List all the different circular permutations of the letters ABCDE. This will yield 24 possibilities. I will leave this one to you. Have fun!! J Please note I will not torture you like this on an exam. 2. How many circular permutations are possible when seating 6 people around a table? Explain. 6P6/6 = 120 The 6P6 represents the number of permutations, but for each permutation there will be 6 repetitions because you can rotate around the circle 6 times (there are 6seats) without changing who is on the left and who is on the right of each person. 3. How many circular combinations are possible when seating four people around a table? Explain. 4C4 = 1 because there is only way to seat the four people around the table. Remember order doesn’t matter. 4. Explain why there are less circular permutations than linear permutations. In circular permutations we have repetitions because we can rotate without changing who is on the person’s left and right. This will not happen with linear permutations. 5. Find the number of circular permutations if I want to create a group of 3 from 4 students. 4P3 / 3 = 8 6. Find the number of circular permutations if I want to create a group of 5 from 10 students. 10P5 / 5 = 6048 7. Find the number of circular permutations if I want to create a group of 7 from 12 students. 12P7 / 7 = 570240 8. Explain why there is no such thing as circular combinations. This question is a little misleading. We simply don’t talk about circular combinations because already order does not matter so the fact that there is no start or end doesn’t change that. In other words there is no difference between linear and circular combinations. 9. Why when I divide by r when calculating a circular permutation do I get rid of all the repeats? r represents the number of seats in the circle. Each permutation can rotate around the table 6 times without changing the person sitting on their left or right. This means there are r repeats for each permutation, and we divide so we get rid of the repetitions for every permutation.
  7. 7. Ms. Jaffe Precalculus Homework due Friday 9/12 (11B, 304), Monday (11A) NAME: _____________________________________________________ 1. What are all the different ways in which combinations / permutations can be used (linear or circular) Ignore this question!! 2. Why does the formula for a circular permutation become nPr if a chair is assigned as the head of the table? The chair at the head of the table created a start and end to the permutation, making it linear. Now each rotation is different because the person at the head of the table changes so there are no repetitions to get rid of. 3. What happens to the formulaif we assign a person to be at the head of the table? Why? The new formula is n-1Pr-1. The permutation is now linear because we have a head of the table. In addition the seat at the head of the table is taken by the person we assign to it so we have one less seat available and one less person to sit. 4. We have a circular table for 4 and 8 people to choose from. If we must choose Michael to sit at the head of the table, in how many ways can we fill the remaining 3 seats? (Michael is one of the 8 people) 7P3 = 210 5. We are trying to create table arrangements for a party with 116 people. Each circular table seats 8 people. How many possible table arrangements are there? You can look at the formula from the Do Now for this day. 116 P P P P P 8 * 108 8 * 100 8 * ... * 12 8  116 14112 8 8 8 8 8 . Note there will be 4 people who are left without seats.
  8. 8. Ms. Jaffe Precalculus Homework due Monday 9/15 (11B) NAME: _________________________________________________________ 1. Explain the process of finding all the permutations of the letters in the word STRESSED. (Make sure your explanation is complete) There are two options. 8! Option 1: 2!3! = 3360. There are 8! permutations of 8 letters. For each permutation there are 2! ways to arrange the Es because there are2 spots dedicated to the Es and 2 Es and 3! ways to arrange the Ss. We need to get rid of all the repetitions so we must divide by 3!2!. Option 2: 8C3 * 5C2 * 3C1 * 2C1 * 1C1 = 3360. The first combination represents the number of ways we can choose three slots in which to place the Ss. We are using combinations because the order in which the put the Ss into the three slots once they are chosen does not matter. The next combination represents the number of ways we can put choose two slots in which to put the Es from the remaining 5. The remaining three combinations tell us the number of ways we can choose slots for each of the three remain letters. Please note that the order in which we choose ing to place the letters is irrelevant so we could have calculated the number of ways to choose 2 slots from 8 in order to put the Es in first instead of the Ss. 2. I have a table for ten, but I only need to seat 6 people. Explain how we find the number of ways in which we can seat these six people. Let’s assume I meant circular since we are discussing circular permutations. Again we have two options. Option 1: Ignore the empty seats in which case it is as if we have a table for 6 and we are seating 6 people = 6P6 / 6 = 120 Option 2: We can take into account the empty seats and treat this is the number of arrangements of 10 letters where four of them are repeats because all the empty seats are the same. In linear permutations this would be 10!/4!, but we are talking about circular so we have to consider that each rotation around the table Is a repetitive permutation. To account for this we must divide by 10 as well. Our final answer is 10!/ (4!*10) = 15120. Notice that ignoring empty seats yields an extremely inaccurate answer. 3. I have 12 people, but a circular table that only seats 8. How many combinations are there? 12C8 = 495. Linear and circular combinations are the same because order is irrelevant. 4. In what cases will a circular permutation become a linear permutation and why? A circular permutation will become a linear permutation as soon as a head of the table is assigned because this creates a start to our permutation and each rotation is now different. 5. I am seating 6 people in a line with 6 seats. If order matters, in how many possible ways can
  9. 9. these 6 people sit if person A and person B cannot sit together? What if they are sitting in a circle? We can calculate the number of permutations in which A and B are sitting together and then subtract that from the total number of permutations which will give us the opposite (the number of permutations in which A and B are not sitting together. To find the number of ways in which A and B are sitting together we should treat A and B as if they are taking up one seat. So there is one less seat and one less pers on. Linear: 5P5 = 120 Circular: 5P5 / 5 = 24 We then multiply each of these numbers by 2 because A could be sitting on Bs left or on Bs right. So the final answers are. Linear: 6P6 – 2(5P5) = 480 Circular: 6P6 / 6 – 2(5P5 / 5) = 120 – 48 = 72 6. What is the probability that if I toss a coin 3 times I get exactly 2 heads? Let us take one possibility: HHT = .5 * .5 * .5 = .125 For each possible combination of Hs and Ts the probability of that combination is .125; so now we only have to find the number of combinations in which we have 2 Hs. It is a combination problem. We are choosing 2 slots from the 3 for Hs or 3C2. We can multiply 3C2 * .125 = .375. 7. Expand (x + y)3 (Note: (x+y)2 = x2 + 2xy + y2) (x+y)(x+y)(x+y)= (x2+2xy+y2)(x+y) = x3 + x2y + 2x2y + 2xy2 + y2x + y3 = x3 + 3x2y + 3xy2 + y3 We are multiplying each term in x2 + 2xy + y2 by x and then by y.
  10. 10. Ms. Jaffe Precalculus Homework due Tuesday 9/16 (304) NAME: __________________________________________________________ 1. Explain the process of finding all the permutations of the letters in the word STRESSED. (Make sure your explanation is complete) There are two options. 8! Option 1: 2!3! = 3360. There are 8! permutations of 8 letters. For each permutation there are 2! ways to arrange the Es because there are2 spots dedicated to the Es and 2 Es and 3! ways to arrange the Ss. We need to get rid of all the repetitions so we must divide by 3!2!. Option 2: 8C3 * 5C2 * 3C1 * 2C1 * 1C1 = 3360. The first combination represents the number of ways we can choose three slots in which to place the Ss. We are using combinations because the order in which the put the Ss into the three slots once they are chosen does not matter. The next combination represents the number of ways we can put choose two slots in which to put the Es from the remaining 5. The remaining three combinations tell us the number of ways we can choose slots for each of the three remain letters. Please note that the order in which we choose ing to place the letters is irrelevant so we could have calculated the number of ways to choose 2 slots from 8 in order to put the Es in first instead of the Ss. 2. If P(A) = .2, P(B) = .4 and P(A and B) = .9, are A and B independent? These are not independent because P(A)P(B)= .2 * .4 = .8  P ( A  B )  .9 3. I have a circular table for ten, but I only need to seat 6 people. Explain how we find the number of ways in which we can seat these six people. Let’s assume I meant circular since we are discussing circular permutations. Again we have two options. Option 1: Ignore the empty seats in which case it is as if we have a table for 6 and we are seating 6 people = 6P6 / 6 = 120 Option 2: We can take into account the empty seats and treat this is the number of arrangements of 10 letters where four of them are repeats because all the empty seats are the same. In linear permutations this would be 10!/4!, but we are talking about circular so we have to consider that each rotation around the table Is a repetitive permutation. To account for this we must divide by 10 as well. Our final answer is 10!/ (4!*10) = 15120. Notice that ignoring empty seats yields an extremely inaccurate answer. 4. I have 12 people, but a circular table that only seats 8. How many combinations are there? 12C8 = 495. Linear and circular combinations are the same because order is irrelevant. 5. In what cases will a circular permutation become a linear permutation and why? A circular permutation will become a linear permutation as soon as a head of the table is assigned because this creates a start to our permutation and each rotation is now different.
  11. 11. 6. I am seating 6 people in a line with 6 seats. If order matters, in how many possible ways can these 6 people sit if person A and person B cannot sit together? What if they are sitting in a circle? We can calculate the number of permutations in which A and B are sitting together and then subtract that from the total number of permutations which will give us the opposite (the number of permutations in which A and B are not sitting together. To find the number of ways in which A and B are sitting together we should treat A and B as if they are taking up one seat. So there is one less seat and one less pers on. Linear: 5P5 = 120 Circular: 5P5 / 5 = 24 We then multiply each of these numbers by 2 because A could be sitting on Bs left or on Bs right. So the final answers are. Linear: 6P6 – 2(5P5) = 480 Circular: 6P6 / 6 – 2(5P5 / 5) = 120 – 48 = 72 7. What is the probability that if I toss a coin 3 times I get exactly 2 heads? Let us take one possibility: HHT = .5 * .5 * .5 = .125 For each possible combination of Hs and Ts the probability of that combination is .125; so now we only have to find the number of combinations in which we have 2 Hs. It is a combination problem. We are choosing 2 slots from the 3 for Hs or 3C2. We can multiply 3C2 * .125 = .375. 8. What is probability of rolling a 2 on the fourth toss of a die? It is 1/6. The events are independent so it is the same as rolling a 2 on any toss of the die. 9. If I designate a certain seat as the head of a table that seats eight people, how many different ways can I seat eight people? 8P8 = 40320. We have created a start. 10. How many circular permutations are there of the letters EIGHTY? 6P6 / 6 = 120 11. Explain why 5C3 = 5C2. The number of ways of choosing 3 things from 5 is the same as the number of ways of not choosing 2 things from 5. I can tell you the number of girls in the class by giving you the total number of students and the number of boys so you can subtract or I can just give you the number of girls. Either way you get the same number. 12. What is the probability I will get exactly 8 heads in ten tosses of a fair coin? Let us take one possibility: HHHHHHHHTT = . 5*.5*.5*. 5*.5*.5*. 5*.5. 5*.5 = .0009765625 For each possible combination of Hs and Ts the probability of that combination is .0009765625; so now we only have to find the number of combinations in which we have 8 Hs. It is a combination problem. We are choosing 8 slots from the 10 for Hs or 10C8. We can multiply 10C8 * .0009765625 = .0439453125 13. A baseball team is playing a five game series. The probability they will win a game is .8.
  12. 12. What is the probability they will win at least 4 out of the five games? 5C4(.8) (.2) + 5C5(.8) (.2) = .4096 + .32768 = .73728 4 1 5 0
  13. 13. Ms. Jaffe Precalculus Homework due Tuesday 9/16 (11B) NAME: _________________________________________________________ 1. If P(A) = .2, P(B) = .4 and P(A and B) = .9, are A and B independent? These are not independent because P(A)P(B)= .2 * .4 = .8  P ( A  B )  .9 2. What is probability of rolling a 2 on the fourth toss of a die? It is 1/6. The events are independent so it is the same as rolling a 2 on any toss of the die. 3. If I designate a certain seat as the head of a table that seats eight people, how many different ways can I seat eight people? 8P8 = 40320. We have created a start. 4. How many circular permutations are there of the letters EIGHTY? 6P6 / 6 = 120 5. Explain why 5C3 = 5C2. The number of ways of choosing 3 things from 5 is the same as the number of ways of not choosing 2 things from 5. I can tell you the number of girls in the class by giving you the total number of students and the number of boys so you can subtract or I can just give you the number of girls. Either way you get the same number. 6. What is the probability I will get exactly 8 heads in ten tosses of a fair coin? Let us take one possibility: HHHHHHHHTT = . 5*.5*.5*. 5*.5*.5*. 5*.5. 5*.5 = .0009765625 For each possible combination of Hs and Ts the probability of that combination is .0009765625; so now we only have to find the number of combinations in which we have 8 Hs. It is a combination problem. We are choosing 8 slots from the 10 for Hs or 10C8. We can multiply 10C8 * .0009765625 = .0439453125 7. A baseball team is playing a five game series. The probability they will win a game is .8. What is the probability they will win at least 4 out of the five games? 5C4(.8) (.2) + 5C5(.8) (.2) = .4096 + .32768 = .73728 4 1 5 0 8. The probability it will rain on any given day is 26%. In a five day week, what is the probability it will rain on at least two days? This is the same as 1 – the probability it will rain on at most one day. 1 – (5C0(..26)0(.74)5 + 5C1(.26)1(.74)4) = 1 – (.2219 + .38983) = .388275
  14. 14. Ms. Jaffe Precalculus Homework due Tuesday 9/16 (11A) NAME: ________________________________________________________ On a separate sheet of paper 1. Finish 2, 3, and 4 from the HW due today. Look Above 2. What is probability of rolling a 2 on the fourth toss of a die? It is 1/6. The events are independent so it is the same as rolling a 2 on any toss of the die. 3. If I designate a certain seat as the head of a table that seats eight people, how many different ways can I seat eight people? 8P8 = 40320. We have created a start. 4. Explain why 5C3 = 5C2. The number of ways of choosing 3 things from 5 is the same as the number of ways of not choosing 2 things from 5. I can tell you the number of girls in the class by giving you the total number of students and the number of boys so you can subtract or I can just give you the number of girls. Either way you get the same number. 5. What is the probability I will get exactly 8 heads in ten tosses of a fair coin? Let us take one possibility: HHHHHHHHTT = . 5*.5*.5*. 5*.5*.5*. 5*.5. 5*.5 = .0009765625 For each possible combination of Hs and Ts the probability of that combination is .0009765625; so now we only have to find the number of combinations in which we have 8 Hs. It is a combination problem. We are choosing 8 slots from the 10 for Hs or 10C8. We can multiply 10C8 * .0009765625 = .0439453125 6. A baseball team is playing a five game series. The probability they will win a game is .8. What is the probability they will win at least 4 out of the five games? 5C4(.8)4(.2)1 + 5C5(.8)5(.2)0 = .4096 + .32768 = .73728 7. The probability it will rain on any given day is 26%. In a five day week, what is the probability it will rain on at least two days? This is the same as 1 – the probability it will rain on at most one day. 1 – (5C0(..26)0(.74)5 + 5C1(.26)1(.74)4) = 1 – (.2219 + .38983) = .388275 8. I have a black and white spinner. The white is 2/3 of the circle and the black is 1/3 of the circle. If I spin 8 times, what is the probability I will land on white less than 6 times? This is the same as 1 – the probability I will land on white 6 or more times. 1- (8C6(2/3)6(1/3)2 + 8C7(2/3)7(1/3)1 + 8C8(2/3)8(1/3)0)= 1 - (.27313 + .15607 + .03902) = .53178 9. Why do you multiply by a combination when calculating the probability of a Bernoulli event?
  15. 15. The combination tells us the number of ways of getting that many successes out of the total number of trials. 10. Next week the probability you will have a test on any given day is .6, what is the probability you will have a test on either 2 or 3 days? 5C2(.6) (.4) + 5C3(.6) (.4) = .2304 + .3456 = .576 2 3 3 2 11. In any of your six periods during the day there is a .9 probability that you will be given homework. What is the probability you will get homework in no more than two classes? 6C0(.9) (.1) + 6C1(.9) (.1) + 6C2(.9) (.1) = .000001 + .000054 + .01215 = .012205 0 6 1 5 2 4 12. Expand the binomial (3x – y)7 1(3x)7(-y)0 + 7(3x)6(-y)1 + 21(3x)5(-y)2 + 35(3x)4(-y)3 + 35(3x)3(-y)4 + 21(3x)2(-y)5 + 7(3x)1(-y)6 + 1(3x)0(-y)7 = 2187x7 – 5103x6y + 5103x5y2 – 2835x4y3 + 945x3y4 – 189x2y5 – 21xy6 – y7
  16. 16. Ms. Jaffe Precalculus Homework due Wednesday 9/17 (11B), Thursday 9/19 (304) NAME: _____________________________________________________________ 1. I have a black and white spinner. The white is 2/3 of the circle and the black is 1/3 of the circle. If I spin 8 times, what is the probability I will land on white less than 6 times? Less than 6 would be 0, 1, 2, 3, 4, 5 times. It is shorter to do the opposite which is 6 or more (6, 7, or 8 times) 1 – [8C6(1/3)6(2/3)2 + 8C7(1/3)7(2/3)1 + 8C8(1/3)8(2/3)0] = .98 2. Why do you multiply by a combination when calculating the probability of a Bernoulli event? The combination tells us the number of ways r successes can occur in n trials. The prqn-r tells us the probability of one way in which r successes occur so we multiply to find the probability of all ways in which r successes occurs. 3. Next week the probability you will have a test on any given day is .6, what is the probability you will have a test on either 2 or 3 days? Note: there are only 5 days in the week on which you can be given a test. 5C2(.6) (.4) + 5C3(.6) (.4) = .576 2 3 3 2 4. In any of your six periods during the day there is a .9 probability that you will be given homework. What is the probability you will get homework in no more than two classes? 6C0(.9)0(.1)6 + 6C1(.9)1(.1)5 + 6C2(.9)2(.1)4 = .00127 5. What is the difference between at least, exactly, and at most when we calculate binomial distribution problems? At least => greater than or equal to Exactly => equal to At most => less than or equal to 6. How do we know when we are using binomial distribution in a problem? We know we are dealing with binomial distribution when there are exactly two outcomes in the experiment and we are looking at r successes in n trials.
  17. 17. Ms. Jaffe Precalculus Homework due Wednesday 9/17 (11A), Thursday 9/18 (11B), Friday 9/19 (304) NAME: _____________________________________________________________ 1. A farm is growing pea pods. The probability that the crop will grow successfully in any given season is .72. a) How do we now we are using binomial distribution? There are only two outcomes, the pea pods will grow successfully or they will not grow successfully. b) What is the probability that the crop will grow successfully in at least 6 of the next 8 seasons? At least means 6, 7, or 8 8C6(.72) (.28) + 8C7(.72) (.28) + 8C8(.72) (.28) = .603 6 2 7 1 8 0 c) What is the probability the crop will grow successfully in at most 3 of the next 7 seasons? At most 3 means 0, 1, 2, or 3 7C0(.72) (.28) + 7C1(.72) (.28) + 7C2(.72) (.28) + 7C3(.72) (.28) = .102 0 7 1 6 2 5 3 4 d) What is the probability the crop will grow unsuccessfully at most 8 of the next 11 seasons? Since we are looking at unsuccessfully, our p = .28 At most 8 means 0, 1, 2, 3, 4, 5, 6, 7, or 8. It is easier to do the opposite which is 9, 10, or 11 1 - [11C9(.28)9(.72)2 + 11C10(.28)10(.72)1 + 11C11(.28)11(.72)0] = .973 2. Explain each part of the binomial distribution formula: nCrprqn-r. The prqn-r tells us the probability of one case in which we r successes in n trials, where we do pr because p is the probability of one success and r is the number of successes and qn-r because q is the probability of one failure and there are n-r failures. nCr is the number of ways we can have r successes in n trials. If we multiple nCr by prqn-r then we get the probability of all cases in which we have r successes in n trials. 3. Why is each row of Pascal’s triangle a palindrome? It is a palindrome because nCr = nCn-r. This is true becaue choosing r things is the same as not choosing the remainder of things or the things left over. 4. Expand the following binomials and simplify completely:
  18. 18. a) (2x + y)4 1(2x) y + 4(2x)3y1 + 6(2x)2y2 + 4(2x)1y3 + 1(2x)0y4 = 4 0 16x4 + 32x3y + 24x2y2 + 8xy3 + y4 b) (-3x + y2)6 1(-3x) (y2)0 + 6(-3x)5(y2)1 + 15(-3x)4(y2)2 + 20(-3x)3(y2)3 + 15(-3x)2(y2)4 + 6(-3x)1(y2)5 + 1(-3x)0(y2)6 = 6 729x6 – 1458x5y2 + 1215x4y4 – 540x3y6 + 135x2y8 – 18xy10 + y12
  19. 19. Ms. Jaffe Precalculus Homework due Wednesday 9/17 (304) NAME: _____________________________________________________________ 1. The probability it will rain on any given day is 26%. In a five day week, what is the probability it will rain on at least two days? This is the same as 1 – the probability it will rain on at most one day. 1 – (5C0(..26)0(.74)5 + 5C1(.26)1(.74)4) = 1 – (.2219 + .38983) = .388275 2. How do we determine if we are dealing with a binomial distribution? We know we are dealing with binomial distribution when there are exactly two outcomes in the experiment and we are looking at r successes in n trials. 3. What is the probability I will get exactly 8 heads in ten tosses of a fair coin? This problem is from your last HW, but redo it using your knowledge from today’s lesson. 10 C8(.5)8(.5)2 = .044 4. A baseball team is playing a five game series. The probability they will win a game is .8. What is the probability they will win at least 4 out of the five games? 5C4(.8)4(.2)1 + 5C5(.8)5(.2)0 = .4096 + .32768 = .73728 5. How do you determine the number of ways you can have a certain number of successes in a binomial distribution? nCr tells us the number of ways we can choose r spots from n so it is also the number of ways we can have r successes in n trials.
  20. 20. Ms. Jaffe Precalculus Homework due Thursday 9/18 (11A), Friday 9/19 (11B) Expand the following binomials: a) (-x + 3y2)5 1(-x)5(3y2)0 + 5(-x)4(3y2)1 + 10(-x)3(3y2)2 + 10(-x)2(3y2)3 + 5(-x)1(3y2)4 + 1(-x)0(3y2)5 = -x5 + 5(x4)(3y2) + 10(-x3)(9y4) + 10(x2)(27y6) + 5(-x)(81y8) + 243y10 = -x5 + 15x4y2 – 90x3y4 + 270x2y6 – 405xy8 + 243y10 Note: you did not need to simplify this one, but I am giving you the answer if you want to practice. b) (x3 – 4y2)8 1(x3)8(-4y2)0 + 8(x3)7(-4y2)1 + 28(x3)6(-4y2)2 + 56(x3)5(-4y2)3 + 70(x3)4(-4y2)4 + 56(x3)3(-4y2)5 + 28(x3)2(-4y2)6 + 8(x3)1(-4y2)7 + 1(x3)0(-4y2)8 = x24 + 8(x21)(-4y2) + 28(x18)(16y4) + 56(x15)(-64y6) + 70(x12)(256y8) + 56(x9)(-1280y10) + 28(x6)(5120y12) + 8(x3)(-20480y14) + 81920y16 = x24 – 32x21y2 + 448x18y4 - 3584 x15y6 + 17920 x12y8 - 71680 x9y10 + 143360 x6y12 - 163840 x3y14 + 81920y16 Note: you did not need to simplify this one, but I am giving you the answer if you want to practice. c) (-2x – y6)9 1(-2x)9(-y6)0 + 9(-2x)8(-y6)1 + 36(-2x)7(-y6)2 + 84(-2x)6(-y6)3 + 126(-2x)5(-y6)4 + 126(-2x)4(-y6)5 + 84(-2x)3(-y6)6 + 36(-2x)2(-y6)7 + 9(-2x)1(-y6)8 + 1(-2x)0(-y6)9 = -512x9 + 9(256x8)(-y6) + 36(-128x7)(y12) + 84(64x6)(-y18) + 126(-32x5)(y24) + 126(16x4)(-y30) + 84(-8x3)(y36) + 36(4x2)(-y42) +9(-2x)(y48) + (-y54) = -512x9 – 2304x8y6 - 4608x7y12 - 5376x6y18 - 4032x5y24 - 2016x4y30 - 672x3y36 - 144x2y42 - 18xy48 - y54 Note: you did not need to simplify this one, but I am giving you the answer if you want to practice. d) (5x + 7y)4 1(5x)4(7y)0 + 4(5x)3(7y)1 + 6(5x)2(7y)2 + 4(5x)1(7y)3 + 1(5x)0(7y)4 = 625x4 + 4(125x3)(7y) + 6(25x2)(49y2) + 4(5x)(343y3) + 2401y4 = 625x4 + 3500x3y + 7350x2y2 + 6860xy3 + 2401y4 e) (4x2 + 3y2)3 1(4x2)3(3y2)0 + 3(4x2)2(3y2)1 + 3(4x2)1(3y2)2 + 1(4x2)0(3y2)3 = 64x6 + 3(16x4)(3y2) + 3(4x2)(9y4) + 27y6 =
  21. 21. 64x6 + 144x4y2 + 108x2y4 + 27y6
  22. 22. Ms. Jaffe Precalculus Homework due Friday 9/19 (11A), Monday 9/22 (11B) NAME: ____________________________________________________ 1. Expand: a) (x3 + 2y5)2 1(x3)2(2y5)0 + 2(x3)1(2y5)1 + 1(x3)0(2y5)2 = x6 + 4x3y5 + 4y10 b) (x - 6y)3 1(x)3(-6y)0 + 3(x)2(-6y)1 + 3(x)1(-6y)2 + 1(x)0(-6y)3 = x3 -18x2y + 18xy2 – 216y3 c) (-2x + y3)5 1(-2x)5(y)0 + 5(-2x)4(y)1 + 10(-2x)3(y)2 + 10(-2x)2(y)3 + 5(-2x)1(y)4 + 1(-2x)0(y)5 = -32x5 + 5(16x4)(y) + 10(-8x3)(y2) + 10(4x2)(y3) + 5(-2x)(y4) + y5 = -32x5 + 80x4y – 80x3y2 + 40x2y3 – 10xy4 + y5 2. I have fifteen jackets. (I am a big jacket person J.) In how many ways can I hang 8 of my jackets on a linear rack? on a circular rack? Linear: P8 = 259459200 15 15 P8 Circular: 8 = 32432400 3. The Yankees need to win at least 6 of the next 10 games to get into the playoffs. During any given game there is a .7 probability that they will win. What is the probability they get into the playoffs? P(6 games) + P(7 games) + P(8 games) + P(9 games) + P(10 games) Note: It doesn’t make it an easier to find the probability of the opposite and subtract from 1 in this problem. 10C6(.7) (.3) + 10C7(.7) (.3) + 10C8(.7) (.3) + 10C9(.7) (.3) + 10C10(.7) (.3) = .8497 6 4 7 3 8 2 9 1 10 0 4. Create a review sheet with as manyWhy questions as possible. Answer as many as you can. Note: There should be at least five.

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