Probability #14 - Questions and Answers

Question 1 Suppose that you are rolling a six sided dice. Let A-get a 3 What is P(A)? α. 4/6 b. 5/6 c. 1/6 d. 3/6 Answer 1 Sample space of a single rolling dice = {1, 2, 3, 4, 5, 6} Event A = get a 3 is {3} P(A) = 1/6 Now P(A’) = 1 - P(A) = 1 - 1/6 = 5/6 Answer: P(A) = 1/6 Answer 2 Ac in this context means A-complement, or "everything that is NOT in A". So if A = less than 2, then Ac = anything 2 or greater. Out of the 6 possible outcomes, 5 of them {2, 3, 4, 5, 6} are in Ac. So P(Ac) = 5/6. Answer 3 Rule P(A') = 1 - P(A) or you roll one die, you roll two dice. if you roll a 6 sided die, then P(A) = 1/6 and its complement would be P(A') = 5/6 or I guess that notation must mean the probability of the complement of A. So it's 5/6. You will almost always NOT get a number smaller than 2! Question 2 10. What is the probability of these events whenwe randomly select a permutation of the 26 lowercase letters of the English alphabet? a) The first 13 letters of the permutation are in alphabetical order. b) a is the first letter of the permutation and z is the last letter. c) a and z are next to each other in the permutation. d) a and b are not next to each other in the permutation. e) a and z are separated by at least 23 letters in the permutation. f) z precedes both a and b in the permutation. Answer a) The first 13 letters of the permutation are in alphabetical order. probability = (26 choose 13) / 26! = 1 / (26 choose 13) d) a and b are not next to each other in the permutation. We need to select 26 out of 26 letters and we need to determine the number of permutations. P(26, 26) = 26! / (26 - 26)! = 26! / 0! = 26! We could find that 2 * 25! of the 26! permutations will have a and b next to each other and thus 26! - 2 * 25! of the 26! permutations will have a and b not to next to each other. The probability is the number of favorable outcomes divided by the number of possible outcomes. P (a and b not next to each other) = (26! - 225!) / 26! = (26! - 225!) / 26! = 1- 2*25! / 26! = 1 - 2/26 = 12/13 = 0.9231 e) The number of possible permutation of 26 letters if 26!. First find the number of ways to place a and z when any one of a and z is placed at first place in 2 ways then another one of a and z can be places in 2 ways at 25th palce or 26th place. So number of ways to place a and z is 4 ways. Now, when any one of a and z is placed at second place in 2 ways then another one of a and z can be places in 1 ways at 26th place. So, number of ways to place a and z is 2 ways. So, total number of ways to place a and z is 4+2=6 The remaining 24 places can be arranged in 24! ways. So, total number of ways to select a permutation of the 26 letters of the English alphabet in such a way that a and z are separate by at least 23 letters in permutation is 6(24!). So, the probability is 6(24!) / 26! = 6 / (26 * 25) = 6 / 650 = 3/325 = 0.00923076923 Question 3 A bowl contains 7 pennies, 9 nickels, and 4 dimes. Elyse removes one coin at random from the bowl and does not replace it. She then removes a second coin at random. What is the probability that both will be dimes? Answer Step 1 7 + 9 + 4 = 20 coins in total Probability of selecting a dime is 4/20 If there is no replacement then there are only 19 coins left now. If we take a dime out on the first pick there are only 3 left now so the second pick is 3/19 The probability that you will pick two dimes is 4/20 * 3/19 = 12 / 380 = 3/95 Question 4 A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it? Answer Step 1 Prior information Probability of an item produced by Machine A: P(A) = 50% = 0.5 Probability of an item produced by Machine B: P(B) = 30% = 0.3 Probability of an item produced by Machine C: P(C) = 20% = 0.2 Considering additional information The likelihood that the product made by A is flawed is: P(D|A) = 3% = 0.03 The likelihood that the product made by B is flawed is: P(D|B) = 2% = 0.02 The likelihood that the product made by C is flawed is: P(D|C) = 1% = 0.01 Given the item is defective, probability of it being produced by B P(B | D) = (P(B) * P(D|B)) / ((P(B) * P(D|B)) + (P(A) * P(D|A)) + (P(C) * P(D|C))) = (0.3 * 0.02) / ((0.3 * 0.02) + (0.5 * 0.03) + (0.2 * 0.01)) = 0.006 / (0.006 + 0.015 + 0.002) = 0.006 / 0.023 = 0.26 Given the item is defective, probability of it being produced by B = 0.26 Question 5 What is the sample space of rolling a 6-sided die? Answer Step 1 Its {1,2,3,4,5,6} which is actually a set of all the possible outcomes as the definition of sample space specifies. The number of dots on the top face of a six-sided die is the result. Now, the outcome of rolling a die is one of 1, 2, 3, 4, or 6 dots appearing on the topmost face. So experiment here is "Rolling a 6 faced dice" and list of possible outcomes is {1,2,3,4,5,6} Sample space by its definition is list of all the possible outcomes of an experiment. So answer to your question is S = {1, 2, 3, 4, 5, 6}