#$&* course MTH 152 Time: 7:32 PMDate: 7/21 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of these 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. More generally, if you have n flips, there are C(n,r) ways to get r Heads. The value of C(n, r) appears in the n+1 row, as the r+1 entry, of Pascal's triangle. STUDENT COMMENT: I solved this question using the given solution for binomials, it might have been more work, but I’m guessing it’s ok? INSTRUCTOR RESPONSE: Your solution is fine. Make sure you also understand the given solution, which is a reasoning process as opposed to a formula. From your previous work I'm confident you do. Knowing how the formula represents the reasoning process, you can then use the formula with confidence. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay. Just have to learn to recognize where I need to use the FCP…I know how to use them, (them being the FCP, collections, permutations, etc) but learning when to use them, more specifically, how to recognize when to use them. ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: What is the significance of .5^2 * .5 for this question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - .5 is the decimal form of 1/2, showing that there are 1/2 * 1/2 * 1/2 options for getting heads, then the last denotes getting tails. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - &&&&Why would one show the probability of no heads P(TTT) first? Why wouldn’t you calculate P(HTT), P(THT), P(TTH)? &&&&& - Regardless, I can see how that out of 8 flips, getting at least one stands a good chance of 7/8. - confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): - Why does one obtain a number based on multiplying the probability of getting tails, when the question states the chances of getting at least one H?
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Given Solution: There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Nailed it. ------------------------------------------------ Self-critique Rating: Ok. Note my confidence between “…choosing at least 1 H from 3” to “3 H out of 7” `q Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Because one side is denoted as a “success” this means any of the other 5, a non-4 number, will be a failure. - I theorize that 6 sides with 3 rolls equates to 6^3 or 216, then C(3,1) in order to demonstrate how any possible outcomes 1 can be taken from 3, then: - 3/216 = .01388 = 1.4 % confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 3 * 1/6 * 25 / 36 = 75 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): - I note that most of my solution was correct, but close. - I can see that in order to get one success out of 3 tries, it must be represented by 1 success and 2 failures. - The probability of rolling a 4 is thus 1/6 and the probability of rolling a non-4 is 5/6 - Thus 1/6 * 5/6 * 5/6, or 1/6 * (5/6) ^2 &&&looking very familiar at this point for the binomial probability formula&&& - Since they could appear in any particular order (4, non-4, non-4) , (non-4, 4, non-4), etc one must state that C(3,1) * 1/6 * (5/6)^2 - At this point in the solution it becomes clear that C(n,r) = p^r * q*(n - r) will be used. - 3 * 1/6 * 5/6 * 5/6 = 25/72 = 34.7% or 35% chance of rolling 1 success and two failures. ------------------------------------------------ Self-critique Rating: Okay. See the amendment to my solution above. `q Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - So this is a case in which one measures the probability of getting 7/10 correct, of which there are 3 choices of 10 questions. - My book describes a scenario in which if a die is rolled, a specific number indicates a specific choice. - 1 or 2 = ‘A’ 3 or 4 = ‘B’ 5 or 6 = ‘C’ - The probability of rolling either 2 numbers out of 6 sides is thus a 33.3% chance for each choice. (33.3 * 3 = 99.9%) - ******************************* - I note that while the preceding information is correct, it is irrelevant, so I will dub this a “binomial probability” scenario. - To get a correct answer there are 1/3 choices. To get a non-correct answer there are 2/3 choices. - There are seven correct answers in question here, so 1/3 ^ 7 * 2/3 ^ 3 - There are C(10,7) ways to get 7 correct answers out of 10 questions, = 120 - C(n, r) * p^r * q ^ (n - r) C(10,7) * (1/3)^7 * (2/3) ^ (10-7) = .01625 = 16.3 % chance. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: `q Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - P of 0 side effects = C(8,0) - P of 1 side effects = C(8,1) - P of 2 side effects = C(8,2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): - I was confused as to how to approach this question, but I see a pattern with your solution. - I can see why C(8,0) * .7 ^ 8 shows the probability of 0 side effects, as opposed to the probability of 4 patients having the side effect as C(8,4) * .7^4 * .3^4. - My main inquiry is regarding the values of p and q represented as .7 and .3, respectively. - Is this by chance a relationship between .7 = .10 - .3, being the percent of patients who endure no side effects? ------------------------------------------------ Self-critique Rating:
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Given Solution: `aThe fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. ** `q Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. ** STUDENT QUESTION: Do we find the 8 steps South and 2 steps North by trial and error? INSTUCTOR RESPONSE: You have to take 10 steps, each north or south. If n stands for the number of steps north and s for the number of steps south, then n and s add up to 10, while the net number of steps south is s - n. We could even up the system of equations n + s = 10 s - n = 6. and solve for n and s. However if you think about the situation, the answer is fairly obvious, so I didn't complicate the solution with the details of this reasoning. Instead I just made the assertion 'To end up 6 blocks South requires 8 steps South and 2 steps North.' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query Add comments on any surprises or insights you experienced as a result of this assignment - I enjoyed the new method of using the binomial probability formula. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Query Add comments on any surprises or insights you experienced as a result of this assignment - I enjoyed the new method of using the binomial probability formula. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!