course Mth 152 I took the first test on Monday and I haven't seen the grade. If possible, could you let me know how I've done? Thanks 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: I understood this. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. What are the odds that the total on a toss of two dice will be greater than 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 of the outcomes will be 4 or less. There are 30 outcomes which total higher than 4. The odds being greater than 4 will be 30:6 or reducing that to 5:1 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: I understood this but in the answer part it says the odds are greater than 6?
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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: I had this backwards in the problem. The probability is because you have to divide what you wish to choose by the whole amount of total objects? ------------------------------------------------ Self-critique rating:2 ********************************************* 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: In a deck of cards there are 52C5 amount of possible hands. In order to get a pair, there are 4C2 ways within the 13 different types of cards. 48 choices for the first of the remaining 3 cards, 44 for the second and 40 for the third (ruling out the possibility of another pair or ruining the pair by having 3 of the same card) These can be chosen in 3 ways in any order, so you have 48*44*40/6 which means out of the 13 types of cards you have 13* 4C2* (48*44*40/6) for one pair in a hand. Use the first part of this and divide it by the total amount of hands possible 52C5. So to find the answer, you must multiply (13*6 or (4C2)*48*44*40/6)/ (52C5) or 52*51*50*49*48/5! 1098240/2598960 which is .42 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: I understood this fairly well, it’s just a lot of steps to work through to arrive at the correct answer so it’s easy to get confused and mess up along the way. Thanks for including the longer explanation for review on how to do it properly and in order. ------------------------------------------------ Self-critique rating:2 ********************************************* 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: If a coin is tossed five times, there are 2*2*2*2*2 or 32 possible outcomes (heads or tails as choices, coin is flipped 5 times for 2^5) For 3 heads, you have 5C3 or 5!*3!/2 20/2=10 outcomes for 3 heads so the probability of 3 heads is 10 outcomes /32 total outcomes reducing to 5/16 or .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. Self-critique: I understood this as well. But just to make sure, getting the 5C3 causes you to 5!*3! And divide by 2 because of the 2 choices for 5*4/2?