#$&* course MTH 152 2/28 12 AM 007. Introduction to probability*********************************************
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Given Solution: There are 6 possible outcomes on the first die and 6 on the second. The number of possible outcomes is therefore 6 * 6 = 36 (e.g., (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), etc.). The six outcomes (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) are all the possible outcomes which are 4 or less. The remaining 36 - 6 = 30 outcomes are all greater than 4. It follows that the probability of obtaining a result greater than 4 is 30 / 36 = 5/6 or .833... or 83.33... %. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. What are the odds that the total on a toss of two dice will be greater than 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 30 outcomes favor the toss of two dice that will be greater than 4, and 6 outcomes do not favor the toss of two dice that will be greater than 4. Therefore, the odds of a toss of two dice being greater than 4 are 30 to 6, or 5 to 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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: There are 15 possible outcomes for the box containing the the balls, 26 possible outcomes for the the box containing the letter tiles, and 7 possible outcomes for the box containing the rings. This means that there are 15*26*7, or 2730, possible choices overall. 8 of the 15 balls are odd, 21 of the 26 letter tiles are consonants, and 3 of the 7 rings are “blue-type” rings. This gives us 8*21*3, or 504, possible choices that contain an odd number, a consonant, and a “blue-type” color. That means that there is a probability of 504 to 2730, or 8*21*3 to 15*26*7 that our collection will contain an odd number, a consonant, and a “blue-type” color. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 15 * 26 * 7 possibilities for the collection obtained by choosing one item from each box. There are 8 odd numbers, 21 consonants if we include 'y' and three 'blue-type' colors. So there are 8 * 21 * 3 possible combinations consisting of an odd number, a consonant and a 'blue-type' color. The desired probability is therefore ( 8 * 21 * 3) / ( 15 * 26 * 7). Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. How many possible 5-card hands can be dealt from a 52-card deck? How many of these hands contain exactly one pair? What therefore is the probability that a hand dealt from a well-shuffled deck will contain exactly one pair? What are the odds in favor of such a deal resulting in a hand with exactly one pair? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 52 cards in a deck, so there are C(52,5) possible 5-card hands. There are 2 cards to a pair, 4 suits, and 13 denominations, so with 2 cards out of 4 suits and 13 possible denominations, we have a possible equation of 4!/(4-2)!*13. The other cards in our 5-card hand cannot be one of the other two cards that match the pair we have chosen, so we have 48 possible cards left to choose from for our 3rd card, 44 possible cards left to choose from for our 4th card, and 40 possible cards left to choose from for our 5th card. This gives us a final equation of 4!/(4-2)!*13*(48*44*40/3!). That means the probability of a hand containing exactly one pair is C(4,2)*13*(48*44*40/3!) to C(52,5). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: There are 2 possible outcomes for each of the 5 tosses, so 2*2*2*2*2, or 32, overall possible outcomes. Using an equation of C(5,3)we find that of 5 flips, there are 10 possible outcomes that we will result in three heads. That means the probability of getting exactly three heads out of five tosses is 10 to 32, or 5 to 16. 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 possibilities 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 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!