Assignment 1 - 08_22_21

MSIS 5503 – STATISTICS FOR DATA SCIENCE ASSIGNMENT 1 – EVENTS AND PROBABILITIES KODJO OPOKU BOTCHWAY A20338464 Q1. a. Sample space of the outcomes {0, 00, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36} = 38 outcomes b. P(red) Outcomes of red = {1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36} Outcomes of red = 18 P(red) = 18 / 38 c. P(-1 st 12-) Outcomes of -1 st 12 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Outcomes of -1 st 12 = 12 P(Outcomes of -1 st 12) = 12 / 38 d. P(Even Number) Even Number = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36} Even Number = 18 P(Even Number) = 18 / 38 e. Getting an odd number is not the not the complement of getting an even number because there exists the presence of 0 and 00 which would fall under the complement of both numbers but are not outcomes of those said sample spaces. That is, the complement of even numbers would include 0 and 00 which are not odd numbers. f. Two mutually exclusive events would be 1. P(Red) and P(Black) 2. P(Even Number) and P(Odd Number) g. The events Even and 1 st Dozen are independent of each other because P(Even ∩ 1 st Dozen) = P(Even) * P(1 st Dozen) 6 / 38 = (18 / 38) * (12 / 38) Q2. Considering the given scenarios, P(C) = 0.4 P(D) = 0.5 P(C|D) = 0.6

a. P(C AND D) The expression P(C AND D) would be represented as P(C∩D) P(C|D) = P(C∩D) / P(D) therefore, P(C∩D) = P(C|D) * P(D) P(C∩D) = 0.6 * 0.5 P(C∩D) = 0.3 b. C and D are not mutually exclusive events because for them to be mutually exclusive then P(C∩D) must be equal to zero. Since it is equal to 0.3, they are not mutually exclusive. c. C and D would be independent events if P(C) is equal to P(C|D). Since they are not equal, they are not independent events. d. P(C OR D) P(C OR D) represented as P(CUD) P(CUD) = P(C) + P(D) -P(C∩D) P(CUD) = 0.4 + 0.5 – 0.3 P(CUD) = 0.6 e. P(D|C) P(D|C) = P(C∩D) / P(C) P(D|C) = 0.3 / 0.4 P(D|C) = 0.75 Q3. a. Table HAIR TYPE BROWN BLOND BLACK RED TOTALS Wavy (W) 20 5 15 3 43 Straight (S) 80 15 65 12 172 Totals 100 20 80 15 215 b. Probability randomly selected child will have wavy hair = Number of children that have wavy hair / Total number of children = 43 / 215 = 0.2 c. Probability randomly selected child will have either brown or blond hair = Number of children that have brown or blond hair / Total number of children = (100 + 20) / 215 = 0.558 d. Probability randomly selected child will have wavy brown hair = Number of children that have wavy brown hair / Total number of children = 20 / 215 = 0.093 e. Probability randomly selected child will have red hair given that he or she has straight hair Number of children that have red and straight hair / Total number of children that have straight hair = 12 / 172 = 0.07 f. B being the event of the child having brown hair

Q5. a. COLLEGE Research Papers (R) No Research Papers (R C ) Total Final Exams (F) 0.32 0.40 0.72 No Final Exams (F C ) 0.14 0.14 0.28 Total 0.46 0.54 1 b. Probability that a course has a final exam or a research project = 1 – P(F C ∩P C ) = 1 – 0.14 = 0.86 c. Find the probability that a course has NEITHER of these two requirements. = P(F C ∩P C ) = 0.14