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Chapter 12: Test Your Proficiency Directions: •Select a section to work on. •Work out each problem on a piece of paper. •Click to check your answer. •For detailed steps click on the provided link. •Move on to the next problem or return to the menu. Sections 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations 12-3, 12-4, 12-5: Probability and Odds 12-6: Statistical Measures 12-7: The Normal Distribution 12-8: Binomial Experiments 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 1. How many ways can you arrange 8 songs on a CD from a list of 13 songs? Check Your Answer Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 1. How many ways can you arrange 8 songs on a CD from a list of 13 songs? There are 51,891,840 ways to arrange the 8 songs. Return to Menu See a detailed explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 1. How many ways can you arrange 8 songs on a CD from a list of 13 songs? Step 1: Determine whether the problem is a Permutation or Combination The word “arrange” means to decide on an order, so this problem is a Permutation. Step 2: Recall the notation and formula for a Permutation n! P n , r where n is the total and r is how many are being taken n r ! Step 3: Determine the values for n and r and plug them into the formula Here n = 13 because there are a total of 13 songs and r = 8 because that’s how many are being taken from the group and arranged Return to Menu 13! 13! P 13,8 51,891,840 13 8! 5! There are 51,891,840 ways to arrange the 8 songs. 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 2. How many 5 card hands can be made that include 4 diamonds and 1 club? Check Your Answer Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 2. How many 5 card hands can be made that include 4 diamonds and 1 club? There are 9295 different 5-card hands with 4 diamonds and 1 club. Return to Menu See a detailed explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 2. How many 5 card hands can be made that include 4 diamonds and 1 club? Step 1: Determine whether the problem is a Permutation or Combination Since the order of the cards doesn’t matter, this is a combination problem. However, because it asks for two different types of cards, it is the product of two separate combinations (two events). Step 2: Recall the notation and formula for a Combination n! C n , r where n is the total and r is how many are being taken n r ! r ! Step 3: Determine the values for n and r for each separate combination and plug them in. For the 4 diamonds: n = 13 (the total number of diamonds) and r = 4 For the 1 club: n = 13 (the total number of clubs) and r = 1 Return to Menu C 13,4 C 13,1 4 diamonds 1 club See the rest of the explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem Step 4: Simplify and state the conclusion. C 13,4 C 13,1 4 diamonds 1 club 13! 13! 9295 9! 4! 12! 1! There are 9295 different 5-card hands with 4 diamonds and 1 club. Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 3. An ice cream shop offers two types of cones, 15 flavors of ice cream and 6 toppings. How many one-scoop cones with two different toppings can be made? Check Your Answer Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 3. An ice cream shop offers two types of cones, 15 flavors of ice cream and 6 toppings. How many one-scoop cones with two different toppings can be made? There are 900 different cones that can be made. Return to Menu See a detailed explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 3. An ice cream shop offers two types of cones, 15 flavors of ice cream and 6 toppings. How many one-scoop cones with two different toppings can be made? Step 1: Determine the number of events. There are four events because four decisions have to be made: type of cone, flavor of ice cream, first topping, second topping Step 2: Determine the number of choices for each event (decision) and use the Fundamental Counting Principle. 2 15 6 5 900 cone flavor 1st topping 2nd topping Step 3: State the conclusion Return to Menu There are 900 different cones that can be made. 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 4. How many ways can the letters in the word INTERCEPT be arranged? Check Your Answer Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 4. How many ways can the letters in the word INTERCEPT be arranged? There are 90,720 ways to arrange the letters. Return to Menu See a detailed explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 4. How many ways can the letters in the word INTERCEPT be arranged? Step 1: Identify the problem as a Permutation with Repetitions. The word “arranged” means the letters are being put in an order, making this problem a permutation. Since some of the letters look exactly the same the problem becomes a permutation with repetitions. Step 2: Recall the formula. n! where n is the total number of letters; p ! q ! p and q represent the number of repetitions of letters that occur more than once Note: If only one object (letter) has repetitions, there will only be a value for p. If more than two objects have repetitions you must divide by additional factorials beyond p and q. Return to Menu See the rest of the explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 4. How many ways can the letters in the word INTERCEPT be arranged? Step 3: Identify the values for n, p, q, etc. as needed. n = 9 because there are 9 letters in the word p = 2 because there are 2 t’s q = 2 because there are 2 q’s Step 4: Plug the values into the formula. 9! 90,720 2 ! 2 ! Step 5: State the conclusion. There are 90,720 ways to arrange the letters. Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 5. How many ways can you choose 9 books from a reading list of 12 books? Check Your Answer Return to Menu 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 5. How many ways can you choose 9 books from a reading list of 12 books? There are 220 ways to choose the 9 books. Return to Menu See a detailed explanation 12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations Go on to the next problem 5. How many ways can you choose 9 books from a reading list of 12 books? Step 1: Identify the problem as a combination. This problem is a combination because no order or position is being assigned to the chosen books. Step 2: Recall the formula for a combination. n! C n , r where n is the total number of objects and n r ! r ! r is how many are being chosen Step 3: Identify the values for n and r and plug them into the formula. n = 12 because there are a total of 12 books r = 9 because that is how many books are being chosen C 12, 9 Return to Menu 12 ! 12! 220 12 9 ! 9 ! 3! 9! There are 220 ways to choose the 9 books. 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 1. A card is drawn from a standard deck. Find the probability the card is a seven or a heart. Check Your Answer Return to Menu 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 1. A card is drawn from a standard deck. Find the probability the card is a seven or a heart. 4 The probability is . 13 Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 1. A card is drawn from a standard deck. Find the probability the card is a seven or a heart. Step 1: Set up the problem. 1) The use of the word “or” means we have to determine whether the events are mutually exclusive or inclusive. P(a seven or a heart) 2) Since the card drawn can be both a seven and a heart at the same time, these events are inclusive. Step 2: Recall the set up for the probability of inclusive events. P A or B P A P B P both A and B P seven or heart P seven P heart P both seven and heart Step 3: Determine each probability and simplify. P seven or heart Return to Menu 4 13 1 16 4 52 52 52 52 13 4 The probability is . 13 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 2. Two cards are drawn in succession from a standard deck without replacement. Find the probability the cards are both hearts. Check Your Answer Return to Menu Go on to the next problem 12-3, 12-4, 12-5: Probability and Odds 2. Two cards are drawn in succession from a standard deck without replacement. Find the probability the cards are both hearts. 1 The probability is . 17 Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 2. Two cards are drawn in succession from a standard deck without replacement. Find the probability the cards are both hearts. Step 1: Set up the problem. P(a heart, then a heart) 1) The use of a comma or the word “and” means we have to determine whether the events are independent or dependent. 2) Since the first card drawn is not being replaced the deck has changed; therefore these events are dependent. Step 2: Recall the set up for the probability of dependent events. P A and B P A P B following A P heart, then heart P heart P heart following heart Step 3: Determine each probability and simplify. P heart, then heart Return to Menu 13 12 1 52 51 17 1 The probability is . 17 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 3. A card is drawn from a standard deck. Find P(red card or face card) Check Your Answer Return to Menu Go on to the next problem 12-3, 12-4, 12-5: Probability and Odds 3. A card is drawn from a standard deck. Find P(red card or face card) 8 The probability is . 13 Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 3. A card is drawn from a standard deck. Find P(red card or face card) Step 1: Set up the problem. P(red card or face card) 1) The use of the word “or” means we have to determine whether the events are mutually exclusive or inclusive. 2) Since the card drawn can be both red and a face card at the same time, these events are inclusive. Step 2: Recall the set up for the probability of inclusive events. P A or B P A P B P both A and B P red or face card P red P face card P both red and face card Step 3: Determine each probability and simplify. P red or face card Return to Menu 26 12 6 32 8 52 52 52 52 13 8 The probability is . 13 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 4. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(4 contemporary and 2 classical). Check Your Answer Return to Menu Go on to the next problem 12-3, 12-4, 12-5: Probability and Odds 4. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(4 contemporary and 2 classical). 108 The probability is . 1001 Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 4. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(4 contemporary and 2 classical). Step 1: Set up the problem. 1) The absence of the word “then” means the CDs are being taken at the same time, therefore order is not important. Since more than one CD is being selected at the same time, combinations must be used. 2) The use of the word “and” means we need to multiply the number of contemporary combinations by the number of classical combinations, and then put this over the total number of ways to select 6 CDs. P 4 contemporary and 2 classical Return to Menu # ways to select 4 contemporary # ways to select 2 classical # ways to select any 6 CDs C 6, 4 C 9,2 15 36 108 C 15,6 5005 1001 108 The probability is . 1001 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 5. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(5 contemporary or 3 classical). Check Your Answer Return to Menu Go on to the next problem 12-3, 12-4, 12-5: Probability and Odds 5. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(5 contemporary or 3 classical). 1734 The probability is . 5005 Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 5. Michelle has a collection of 15 CDs: 6 contemporary and 9 classical. If she selects 6 of the CDs at random, find: P(5 contemporary or 3 classical). Step 1: Identify the characteristics of the problem. 1) 2) 3) In this problem, the use of the word “or” separates the groups of CDs being chosen making each group less than the total number of 6 being chosen. It also means we will add. We need to make up the total of 6 for each group: 5 contemporary becomes 5 contemporary and 1 classical 3 classical becomes 3 classical and 3 contemporary Since more than one CD is being selected at once we will need to use combinations. Step 2: Set up the problem. P 5 contemporary or 3 classical P 5 contemporay and 1 classical or 3 classical and 3 contemporary C 6, 5 C 9, 1 C 9,3 C 6, 3 6 9 84 20 1734 C 15,6 5005 5005 Return to Menu The probability is 1734 . 5005 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 6. The probability an event will happen is Find the odds the event will happen. Check Your Answer Return to Menu 8 . 15 Go on to the next problem 12-3, 12-4, 12-5: Probability and Odds 6. The probability an event will happen is 8 . 15 Find the odds the event will happen. The odds are 8:7. Return to Menu See a detailed explanation 12-3, 12-4, 12-5: Probability and Odds Go on to the next problem 8 6. The probability an event will happen is . 15 Find the odds the event will happen. Step 1: Recall the definitions for probability and odds. Given s = success and f = failure P s s s f and Odds = s : f Step 2: Using the given information to find the values for s and f. 8 s therefore, s 8 and s f 15 15 s f So, 8 f 15 f 7 P s Return to Menu Step 3: Use the definition of odds to answer the problem. Odds = s : f 8 : 7 The odds are 8:7. 12-6: Statistical Measures Go on to the next problem 1. In his last eight basketball games Jeremy scored the following number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean, median, and mode for this set of data. Check Your Answer Return to Menu 12-6: Statistical Measures Go on to the next problem 1. In his last eight basketball games Jeremy scored the following number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean, median, and mode for this set of data. The mean is 14.375, the median is 15 and the mode is 17. Return to Menu See a detailed explanation 12-6: Statistical Measures Go on to the next problem 1. In his last eight basketball games Jeremy scored the following number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean, median, and mode for this set of data. Step 1: Recall the definitions for mean, median, and mode. Mean = the sum of the data values divided by the number of data values Median = the middle value when the data are in numerical order Mode = the data value(s) with the greatest frequency Step 2: Apply the definitions to find each value. Put the data in order: 6, 8, 10, 13, 17, 17, 20, 24 115 Mean = 14.375 8 13+17 Median = 15; here you must average the two middle values 2 Mode = 17 because there are two 17s and only one of everything else Return to Menu The mean is 14.375, the median is 15 and the mode is 17. 12-6: Statistical Measures Go on to the next problem 2. Find the variance and standard deviation for the set of data {28, 31, 15, 44, 22}. Round to the nearest tenth if necessary. Check Your Answer Return to Menu 12-6: Statistical Measures Go on to the next problem 2. Find the variance and standard deviation for the set of data {28, 31, 15, 44, 22}. Round to the nearest tenth if necessary. The variance is 94 and the standard deviation is about 9.7. Return to Menu See a detailed explanation 12-6: Statistical Measures Go on to the next problem 2. Find the variance and standard deviation for the set of data {28, 31, 15, 44, 22}. Round to the nearest tenth if necessary. Step 1: Recall the steps for finding the variance. 1) 2) 3) 4) Find the mean Find the difference between the mean and each data value Square the differences Find the mean of the squares Step 2: Follow the steps to find the variance. step 1) Find the mean: steps 2 4) variance Return to Menu 28 31 15 44 22 28 5 2 2 2 2 2 28 28 31 28 28 15 44 28 28 22 5 02 32 132 162 62 470 94 5 5 See the rest of the explanation 12-6: Statistical Measures Go on to the next problem 2. Find the variance and standard deviation for the set of data {28, 31, 15, 44, 22}. Round to the nearest tenth if necessary. Step 3: Recall the definition of the standard deviation. The standard deviation is the square root of the variance. Step 4: Find the standard deviation. standard deviation = 94 9.7 Step 5: State the conclusion. The variance is 94 and the standard deviation is about 9.7. Return to Menu 12-7: The Normal Distribution Go on to the next problem 1. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored between 71 and 95? Check Your Answer Return to Menu 12-7: The Normal Distribution Go on to the next problem 1. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored between 71 and 95? About 49.5% scored between 71 and 95. Return to Menu See a detailed explanation 12-7: The Normal Distribution Go on to the next problem 1. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored between 71 and 95? Step 1: Make a number line using the given mean and standard deviation and the corresponding percentages for a normal distribution. 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 2: Identify whether the answer should be a percentage or whole number. For this problem the phrase “what percent” means the answer is a percentage number. Return to Menu See the rest of the explanation 12-7: The Normal Distribution Go on to the next problem 1. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested? Step 3: Identify the correct intervals. “scored between 71 and 95” 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 4: Add the percentages and state the conclusion. 34% + 13.5% + 2% = 49.5% Return to Menu About 49.5% scored between 71 and 95. 12-7: The Normal Distribution Go on to the next problem 2. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored less than 79? Check Your Answer Return to Menu Go on to the next problem 12-7: The Normal Distribution 2. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored less than 79? About 84% scored less than 79. Return to Menu See a detailed explanation 12-7: The Normal Distribution Go on to the next problem 2. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored less than 79? Step 1: Make a number line using the given mean and standard deviation and the corresponding percentages for a normal distribution. 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 2: Identify whether the answer should be a percentage or whole number. For this problem the phrase “what percent” means the answer is a percentage number. Return to Menu See the rest of the explanation 12-7: The Normal Distribution Go on to the next problem 2. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. What percent of those tested scored less than 79? Step 3: Identify the correct intervals. “scored less than 79” 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 4: Add the percentages and state the conclusion. 0.5% + 2% + 13.5% + 34% + 34% = 84% Return to Menu About 84% scored less than 79. 12-7: The Normal Distribution Go on to the next problem 3. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. How many of those tested scored between 55 and 95? Check Your Answer Return to Menu 12-7: The Normal Distribution Go on to the next problem 3. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. About how many of those tested scored between 55 and 95? About 776 of those tested scored between 55 and 95. Return to Menu See a detailed explanation 12-7: The Normal Distribution Go on to the next problem 3. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. About how many of those tested scored between 55 and 95? Step 1: Make a number line using the given mean and standard deviation and the corresponding percentages for a normal distribution. 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 2: Identify whether the answer should be a percentage or whole number. For this problem the phrase “about how many” means the answer is a whole number. Return to Menu See the rest of the explanation 12-7: The Normal Distribution Go on to the next problem 3. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. About how many of those tested scored between 55 and 95? Step 3: Identify the correct intervals. “scored between 55 and 95” 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 4: 1) Add the percentages. 2) Find this percentage of the total number of scores on the test. 3) State the conclusion. Note: if the Return to Menu 1) 13.5% + 34% + 34% + 13.5% + 2% = 97% 2) 97% of 800 scores = 0.97(800) = 776 answer in part 2 is a decimal, round to the nearest whole number About 776 of those tested scored between 55 and 95. 12-7: The Normal Distribution Go on to the next problem 4. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. About how many of those tested scored more than 79? Check Your Answer Return to Menu 12-7: The Normal Distribution Go on to the next problem 4. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. How many of those tested scored more than 79? About 128 of those tested scored more than 79. Return to Menu See a detailed explanation 12-7: The Normal Distribution Go on to the next problem 4. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. How many of those tested scored more than 79? Step 1: Make a number line using the given mean and standard deviation and the corresponding percentages for a normal distribution. 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 2: Identify whether the answer should be a percentage or whole number. For this problem the phrase “about how many” means the answer is a whole number. Return to Menu See the rest of the explanation 12-7: The Normal Distribution Go on to the next problem 4. The scores on 800 tests are normally distributed with a mean score of 71 and a standard deviation of 8. How many of those tested scored more than 79? Step 3: Identify the correct intervals. “scored more than 79” 0.5% 2% 47 13.5% 55 34% 63 13.5% 34% 71 mean 79 2% 87 0.5% 95 Step 4: 1) Add the percentages. 2) Find this percentage of the total number of scores on the test. 3) State the conclusion. Note: if the 1) 13.5% + 2% + 0.5% = 16% 2) 16% of 800 scores = 0.16(800) = 128 Return to Menu About 128 of those tested scored more than 79. answer in part 2 is a decimal, round to the nearest whole number 12-8: Binomial Experiments Go on to the next problem 1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes exactly 4 goals? Check Your Answer Return to Menu Go on to the next problem 12-8: Binomial Experiments 1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes exactly 4 goals? The probability is Return to Menu 448 . 15,625 See a detailed explanation 12-8: Binomial Experiments Go on to the next problem 1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes exactly 4 goals? Step 1: Define the two possible outcomes and identify the power of the corresponding binomial. Let x = makes a goal Let y = misses a goal The power of the binomial is 7 because Julio is taking 7 shots at the goal. Step 2: Set up the binomial and expand using the binomial theorem. x y 7 x7 7x 6 y 7 6x 5 y 2 7 6 5 x 4 y 3 7 6 5 4 x 3 y 4 0! 1! 2! 3! 4! 7 6 5 4 3 x 2 y 5 7 6 5 4 3 2 xy 6 7! y 7 5! 6! 7! x y 7 x 7 7x 6 y 21x 5 y 2 35x 4 y 3 35x 3 y 4 21x 2 y 5 7xy 6 x 7 Return to Menu See the rest of the explanation 12-8: Binomial Experiments Go on to the next problem 1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes exactly 4 goals? Step 3: Determine which term(s) should be used. “makes exactly 4 goals” 4 3 Use 35x y The term with x4 should be used because x stands for making a goal and we are finding the probability of making 4 goals. Step 4: Find the probabilities for making a goal and missing a goal. 1 (this is given information) 5 1 4 P missing a goal P y 1 5 5 P making a goal P x Step 5: Plug in the probabilities for x and y and simplify. 4 3 35 1 64 448 1 4 Use 35x 4 y 3 35 1 625 125 15,625 5 5 Return to Menu The probability is 448 . 15,625 12-8: Binomial Experiments Go on to the next problem 2. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes 2 or 3 goals? Check Your Answer Return to Menu Go on to the next problem 12-8: Binomial Experiments 2. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes 2 or 3 goals? The probability is Return to Menu 30,464 . 78,125 See a detailed explanation 12-8: Binomial Experiments Go on to the next problem 2. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes 2 or 3 goals? Step 1: Define the two possible outcomes and identify the power of the corresponding binomial. Let x = makes a goal Let y = misses a goal The power of the binomial is 7 because Julio is taking 7 shots at the goal. Step 2: Set up the binomial and expand using the binomial theorem. x y 7 x7 7x 6 y 7 6x 5 y 2 7 6 5 x 4 y 3 7 6 5 4 x 3 y 4 0! 1! 2! 3! 4! 7 6 5 4 3 x 2 y 5 7 6 5 4 3 2 xy 6 7! y 7 5! 6! 7! x y 7 x 7 7x 6 y 21x 5 y 2 35x 4 y 3 35x 3 y 4 21x 2 y 5 7xy 6 x 7 Return to Menu See the rest of the explanation 12-8: Binomial Experiments Go on to the next problem 2. The probability that Julio makes a goal is 1/5. If he takes 7 shots at the goal, what is the probability that he makes 2 or 3 goals? Step 3: Determine which term(s) should be used. “makes 2 or 3 goals” Use 35x 3 y 4 21x 2 y 5 The terms with x2 and x3 should be used because x stands for making a goal and we are finding the probability of making 2 or 3 goals. Step 4: Find the probabilities for making a goal and missing a goal. 1 (this is given information) 5 1 4 P missing a goal P y 1 5 5 P making a goal P x Step 5: Plug in the probabilities for x and y and simplify. 3 4 1 4 1 Use 35x y 21x y 35 21 5 5 5 35 1 256 21 1 1024 30, 464 1 125 625 1 25 3125 78,125 3 4 2 5 2 4 5 5 The probability is Return to Menu 30,464 . 78,125 12-8: Binomial Experiments 3. The probability that Juan makes a goal is 3/5. If he takes 6 shots at the goal, what is the probability that he makes at least 4 goals? Check Your Answer Return to Menu 12-8: Binomial Experiments 3. The probability that Juan makes a goal is 3/5. If he takes 6 shots at the goal, what is the probability that he makes at least 4 goals? The probability is Return to Menu 1701 . 3125 See a detailed explanation 12-8: Binomial Experiments 3. The probability that Juan makes a goal is 3/5. If he takes 6 shots at the goal, what is the probability that he makes at least 4 goals? Step 1: Define the two possible outcomes and identify the power of the corresponding binomial. Let x = makes a goal Let y = misses a goal The power of the binomial is 6 because Juan is taking 6 shots at the goal. Step 2: Set up the binomial and expand using the binomial theorem. x y 6 x6 6x 5 y 6 5x 4 y 2 6 5 4 x 3 y 3 6 5 4 3 x 2 y 4 0! 1! 2! 3! 4! 6 5 4 3 2 x 1y 5 6! y 6 5! 6! x y 6 x 6 6x 5 y 15x 4 y 2 20x 3 y 3 15x 2 y 4 6xy 5 x 6 Return to Menu See the rest of the explanation 12-8: Binomial Experiments 3. The probability that Juan makes a goal is 3/5. If he takes 6 shots at the goal, what is the probability that he makes at least 4 goals? Step 3: Determine which term(s) should be used. “at least 4 goals” Use x 6 6x 5 y 15x 4 y 2 The terms with x6 and x5 and x4 should be used because x stands for making a goal and we are finding the probability of making at least 4 goals (4 or 5 or 6 goals). Step 4: Find the probabilities for making a goal and missing a goal. 3 (this is given information) 5 3 2 P missing a goal P y 1 5 5 P making a goal P x Step 5: Plug in the probabilities for x and y and simplify. 6 5 4 3 3 2 3 2 Use x 6x y 15x y 6 15 5 5 5 5 5 729 6 243 2 15 81 4 8505 1701 15,625 1 3125 5 1 625 25 15,625 3125 6 5 4 2 2 The probability is Return to Menu 1701 . 3125