#$&* course Mth 152 Dec 5 - 5:50pm 009. ``q Query 9
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Given Solution: On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. • Of these 8 possibilities, only 3 of them have two Heads. • Thus the probability is 3 / 8. You can get this result without listing. • There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. • To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. • There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. More generally, if you have n flips, there are C(n,r) ways to get r Heads. The value of C(n, r) appears in the n+1 row, as the r+1 entry, of Pascal's triangle. STUDENT COMMENT: I solved this question using the given solution for binomials, it might have been more work, but I’m guessing it’s ok? INSTRUCTOR RESPONSE: Your solution is fine. Make sure you also understand the given solution, which is a reasoning process as opposed to a formula. From your previous work I'm confident you do. Knowing how the formula represents the reasoning process, you can then use the formula with confidence. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: What is the significance of .5^2 * .5 for this question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .5^2*.5 We saw this formula in the previous problem. .5 = the probability of getting tails .5^2 = probability of heads .5^2*.5 = 0.25*.5 = 0.125 = .375 or 3/8 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): okay - I understand the combination formula route. ------------------------------------------------ Self-critique Rating: okay ********************************************* 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: If three fair coins are tossed, find the probability of each number of heads/at least one 2*2*2= 8 Getting at least one head would be 7/8 probability. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want 3 heads on 7 flips. C (7,3) 7!/3! (7-3)! 7!/3! (4!) 7 * 6 * 5 * 4 * 3 * 2 * 1/ 3*2*1 (4*3*2*1) 7 * 6 * 5 / 3*2*1 210/6 35 ways to have three heads for 7 flips To find the probability we need to know the total other ways of any of the flips. 1/2 ^3 * 1/2 ^4 = 1/ 2^ 7 = 1/ 128 ways 35 * 1/128 = 35/128 = = .2734 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): okay ------------------------------------------------ Self-critique Rating: okay `q Query 12.4.21 A fair die is rolled three times. A 4 is considered ‘success’ while all other outcomes are failures. Find the probability of each number of successes = 1 success. Your solution: We roll a fair die 3 times. To only get one success (4) out of it would be a success of 5/6 since there are 6 sides and successes on a die. Since we are looking for that 1 success specifically on each of three rolls, there will be 3 ways to get that 4. To find the probability we need to multiply 5/6 by the 3 ways by 1/6. C (3,1) * 1/6 * (5/6)^2 = 3 * 1/6 * 25 / 36 = 75 / 216 = 25 / 72 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 3 * 1/6 * 25 / 36 = 75 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How do we get back to 1 success? Self-critique Rating: 1 `q Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 question multiple choice - each question has 3 choices, one is correct. By rolling a fair die she finds the answer she’ll choose. We are looking for exactly seven correct answers. Each choice for each question has a 1/3 chance of being correct. There are 10 questions in all. We are looking for 7 of those to be correct. C (10,7) = 7 ways of getting the correct answers. C (10,7) * (2/3)^3 * (1/3)^7 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 1 `q Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each more than two have undesirable effects. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C (8,2) 8! / 2! (8-2)! 8! / 2! (6!) 8*7*6*5*4*3*2*1/2*1 (6*5*4*3*2*1) 8*7/2*1 56/2 28 C(8,2) * .7^6 * .3^2. 28 * .7^6 * .3^2 28 *.12 * .09 .3024 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m sure if I added wrong but I didn’t get .448 ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `aThe fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay `q Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it? 54. On her corner. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024 C (10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 = 10s - 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: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!