#$&* course Mth 152 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: There are 6 possible outcomes on the first die and 6 on the second. The number of possible outcomes is therefore 6 * 6 = 36 (e.g., (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), etc.). The six outcomes (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) are all the possible outcomes which are 4 or less. The remaining 36 - 6 = 30 outcomes are all greater than 4. It follows that the probability of obtaining a result greater than 4 is 30 / 36 = 5/6 or .833... or 83.33... %. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q002. What are the odds that the total on a toss of two dice will be greater than 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 to 1 odds 5 favorable, 1 unfavorable outcome out of 6 possibilities confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As seen in the previous question, there are 30 possible outcomes or the total is greater than 4 and 6 outcomes where the outcome is less than or equal to 4. The odds in favor of any event are expressed as odds = number in favor to number opposed. {}In this case the odds of a result greater than 6 are 30 to 6, which reduces to 5 to 1. These odds can also be expressed as 30 : 6 or 5 : 1. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q003. Suppose we have three boxes, one containing balls numbered 1-15, another tiles labeled a-z, and another one ring for each of the seven colors of the rainbow. How many possibilities are there for the collection of items we obtain if we choose one item from each box? How many of these possibilities contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? If we choose one item from each box, what is the probability that our collection will contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15*26*7=2730 8*21*3=504 504/2730 .18461 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 15 * 26 * 7 possibilities for the collection obtained by choosing one item from each box. There are 8 odd numbers, 21 consonants if we include 'y' and three 'blue-type' colors. So there are 8 * 21 * 3 possible combinations consisting of an odd number, a consonant and a 'blue-type' color. The desired probability is therefore ( 8 * 21 * 3) / ( 15 * 26 * 7). Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q004. How many possible 5-card hands can be dealt from a 52-card deck? How many of these hands contain exactly one pair? What therefore is the probability that a hand dealt from a well-shuffled deck will contain exactly one pair? What are the odds in favor of such a deal resulting in a hand with exactly one pair? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C[52,5] 52!/5![52-5]! 52!/5!47!=2,598,960 [pg 698, ed 11] there are 1,098,240 pair[pg 701] 1,098,240/2,598,960 .423 probability odds are close to 1:1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(52, 5) possible hands. There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third. Any given combination of the three remaining cards can be chosen in any of 3! ways so there are 48 * 44 * 40 possible choices of these 3 cards. Thus there are 13 * C(4, 2) * ( 48 * 44 * 40 / 3!) hands containing of exactly one pair. The probability of exactly one pair is therefore [ 13 * C(4,2) * (48 * 44 * 40) / 3! ] / C(52,5). This expression is easily enough written out and reduced [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = 6 * 44 * 4 * 5 * 4 / [ 4 * 51 * 5 * 49 ] = 6 * 44 * 4 / [ 51 * 49 ] = (24 * 44) / (51 * 49) = (8 * 44) / ( 17 * 49) = .42 approx. Further explanation: This builds on the ideas of permutations and combinations developed in previous assignments. To get a hand you have to 'choose' 5 of the 52 cards, and order doesn't matter. There are C(52, 5) ways of doing this To get a pair of 5's, for example, you have to choose 2 of the four 5's in the deck. There are C(4, 2) ways to do this. There are 13 denominations (2's, 3's, 4's, ..., Queens, Kings). The pair could be from anyone of these denominations so there are 13 * C(4,2) ways to get a pair. After choosing the pair, you can't choose another card of that denomination or you would no longer have a pair. That leaves only 48 cards from which to choose the third. You already have a pair so the next card can't match the denomination of the third, so you have only 44 cards from which to choose the fourth. Similar reasoning shows that there are only 40 cards from which to choose the fifth card. These last three cards could have been chosen in any of 3! orders. So the number of ways of choosing the last three cards is 48*44*40/3!. So by the fundamental counting principle, since we have to choose a pair and then choose three other cards not matching the denomination of the pair or of one another, the number of possible ways to accomplish this is 13 * C(4,2) * 48 * 44 * 40 / 3!. STUDENT QUESTION I get confused in this step: “There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third.” After you get a pair, I would think you would have 44 choices for the first of the remaining three cards, 40 for the second, and 36 for the third because 52 and 48 were for the pair??? INSTRUCTOR RESPONSE: Having gotten a pair, which consists of two cards of the same denomination, you can't get another card of that denomination (if you did you would no longer have a pair, but at least three of a kind). The next card could be any card not of that denomination, and there are 48 such cards. STUDENT QUESTION: Also, I don’t understand how you simplified the answer. 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = Why did you move 5*4*3*2*1 to the first half of the equation and 3*2*1 to the second half??? INSTRUCTOR RESPONSE: Division by a fraction is the same as multiplication by the reciprocal (for example (a / b) / (c / d) = (a / b) * (d / c) = a * d / (b * c). In this case we are dividing by [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ], so we get [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = [ 13 * 6 * 48 * 44 * 40 / (3 * 2 * 1) ] * [ (5 * 4 * 3 * 2 * 1) / (52 * 51 * 50 * 49 * 48) ]. We now multiply the numerators of the two fractions, and the denominators, to get 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] , and then continue as indicated in the given solution. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q005. If a fair coin is tossed five times, how many possible outcomes are there? How many of these outcomes will have exactly 3 'heads'? What therefore is the probability that on 5 tosses of a fair coin we will obtain exactly 3 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c[5,3] 5!/3![5-3]! 5*4*3*2*1/3![2!] 5*4*3/3*2*1 60/6=10 2*2*2*2*2=32 10/32=5/16 .3125 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: On 5 flips there are C(5,3) = 10 possible outcomes with exactly 3 'heads'. There are 2^5 = 32 possible outcomes altogether. The probability of 3 'heads' on 5 flips is therefore 10 / 32 = 5/16 = .3125. STUDENT QUESTION I don’t understand where you got the 32 from? Did I figure out the possiblilities of 5 tosses wrong because I came up with 80 possible outcomes. INSTRUCTOR RESPONSE On five flips there are 2 possibilities on the first flip, 2 possibilities on the second, 2 possibilities on the third, 2 possibilities on the fourth and 2 possibilities on the fifth. By the Fundamental Counting Principle there are 2 * 2 * 2 * 2 * 2 = 2^5 = 32 possible outcomes on 5 flips. You can also see this by making a 'tree' diagram of the possibilities. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q006. What is the probability that when rolling two dice the total will be at least 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 33/36 .917 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique Ratingok ********************************************* Question: `q007. What is the probability of obtaining a 5-card hand of all red cards, when dealt from a 52-card deck? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c[26,5] 26!/5![26-5] 26!/5![21!] 26*25*24*23*22/5*4*3*2*1 7893600/120=65780 65780/2598960[#of poker hands, pge 701] .0253 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 152 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: There are 6 possible outcomes on the first die and 6 on the second. The number of possible outcomes is therefore 6 * 6 = 36 (e.g., (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), etc.). The six outcomes (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) are all the possible outcomes which are 4 or less. The remaining 36 - 6 = 30 outcomes are all greater than 4. It follows that the probability of obtaining a result greater than 4 is 30 / 36 = 5/6 or .833... or 83.33... %. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q002. What are the odds that the total on a toss of two dice will be greater than 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 to 1 odds 5 favorable, 1 unfavorable outcome out of 6 possibilities confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As seen in the previous question, there are 30 possible outcomes or the total is greater than 4 and 6 outcomes where the outcome is less than or equal to 4. The odds in favor of any event are expressed as odds = number in favor to number opposed. {}In this case the odds of a result greater than 6 are 30 to 6, which reduces to 5 to 1. These odds can also be expressed as 30 : 6 or 5 : 1. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q003. Suppose we have three boxes, one containing balls numbered 1-15, another tiles labeled a-z, and another one ring for each of the seven colors of the rainbow. How many possibilities are there for the collection of items we obtain if we choose one item from each box? How many of these possibilities contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? If we choose one item from each box, what is the probability that our collection will contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15*26*7=2730 8*21*3=504 504/2730 .18461 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 15 * 26 * 7 possibilities for the collection obtained by choosing one item from each box. There are 8 odd numbers, 21 consonants if we include 'y' and three 'blue-type' colors. So there are 8 * 21 * 3 possible combinations consisting of an odd number, a consonant and a 'blue-type' color. The desired probability is therefore ( 8 * 21 * 3) / ( 15 * 26 * 7). Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q004. How many possible 5-card hands can be dealt from a 52-card deck? How many of these hands contain exactly one pair? What therefore is the probability that a hand dealt from a well-shuffled deck will contain exactly one pair? What are the odds in favor of such a deal resulting in a hand with exactly one pair? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C[52,5] 52!/5![52-5]! 52!/5!47!=2,598,960 [pg 698, ed 11] there are 1,098,240 pair[pg 701] 1,098,240/2,598,960 .423 probability odds are close to 1:1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(52, 5) possible hands. There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third. Any given combination of the three remaining cards can be chosen in any of 3! ways so there are 48 * 44 * 40 possible choices of these 3 cards. Thus there are 13 * C(4, 2) * ( 48 * 44 * 40 / 3!) hands containing of exactly one pair. The probability of exactly one pair is therefore [ 13 * C(4,2) * (48 * 44 * 40) / 3! ] / C(52,5). This expression is easily enough written out and reduced [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = 6 * 44 * 4 * 5 * 4 / [ 4 * 51 * 5 * 49 ] = 6 * 44 * 4 / [ 51 * 49 ] = (24 * 44) / (51 * 49) = (8 * 44) / ( 17 * 49) = .42 approx. Further explanation: This builds on the ideas of permutations and combinations developed in previous assignments. To get a hand you have to 'choose' 5 of the 52 cards, and order doesn't matter. There are C(52, 5) ways of doing this To get a pair of 5's, for example, you have to choose 2 of the four 5's in the deck. There are C(4, 2) ways to do this. There are 13 denominations (2's, 3's, 4's, ..., Queens, Kings). The pair could be from anyone of these denominations so there are 13 * C(4,2) ways to get a pair. After choosing the pair, you can't choose another card of that denomination or you would no longer have a pair. That leaves only 48 cards from which to choose the third. You already have a pair so the next card can't match the denomination of the third, so you have only 44 cards from which to choose the fourth. Similar reasoning shows that there are only 40 cards from which to choose the fifth card. These last three cards could have been chosen in any of 3! orders. So the number of ways of choosing the last three cards is 48*44*40/3!. So by the fundamental counting principle, since we have to choose a pair and then choose three other cards not matching the denomination of the pair or of one another, the number of possible ways to accomplish this is 13 * C(4,2) * 48 * 44 * 40 / 3!. STUDENT QUESTION I get confused in this step: “There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third.” After you get a pair, I would think you would have 44 choices for the first of the remaining three cards, 40 for the second, and 36 for the third because 52 and 48 were for the pair??? INSTRUCTOR RESPONSE: Having gotten a pair, which consists of two cards of the same denomination, you can't get another card of that denomination (if you did you would no longer have a pair, but at least three of a kind). The next card could be any card not of that denomination, and there are 48 such cards. STUDENT QUESTION: Also, I don’t understand how you simplified the answer. 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = Why did you move 5*4*3*2*1 to the first half of the equation and 3*2*1 to the second half??? INSTRUCTOR RESPONSE: Division by a fraction is the same as multiplication by the reciprocal (for example (a / b) / (c / d) = (a / b) * (d / c) = a * d / (b * c). In this case we are dividing by [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ], so we get [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = [ 13 * 6 * 48 * 44 * 40 / (3 * 2 * 1) ] * [ (5 * 4 * 3 * 2 * 1) / (52 * 51 * 50 * 49 * 48) ]. We now multiply the numerators of the two fractions, and the denominators, to get 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] , and then continue as indicated in the given solution. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q005. If a fair coin is tossed five times, how many possible outcomes are there? How many of these outcomes will have exactly 3 'heads'? What therefore is the probability that on 5 tosses of a fair coin we will obtain exactly 3 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c[5,3] 5!/3![5-3]! 5*4*3*2*1/3![2!] 5*4*3/3*2*1 60/6=10 2*2*2*2*2=32 10/32=5/16 .3125 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: On 5 flips there are C(5,3) = 10 possible outcomes with exactly 3 'heads'. There are 2^5 = 32 possible outcomes altogether. The probability of 3 'heads' on 5 flips is therefore 10 / 32 = 5/16 = .3125. STUDENT QUESTION I don’t understand where you got the 32 from? Did I figure out the possiblilities of 5 tosses wrong because I came up with 80 possible outcomes. INSTRUCTOR RESPONSE On five flips there are 2 possibilities on the first flip, 2 possibilities on the second, 2 possibilities on the third, 2 possibilities on the fourth and 2 possibilities on the fifth. By the Fundamental Counting Principle there are 2 * 2 * 2 * 2 * 2 = 2^5 = 32 possible outcomes on 5 flips. You can also see this by making a 'tree' diagram of the possibilities. Self-critique:ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q006. What is the probability that when rolling two dice the total will be at least 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 33/36 .917 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique Ratingok ********************************************* Question: `q007. What is the probability of obtaining a 5-card hand of all red cards, when dealt from a 52-card deck? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c[26,5] 26!/5![26-5] 26!/5![21!] 26*25*24*23*22/5*4*3*2*1 7893600/120=65780 65780/2598960[#of poker hands, pge 701] .0253 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!