Probability #1 - Questions and Answers

Question 1 If you used a random number generator for the numbers from 1 through 20 to play a game, what is the theoretical probability of getting each of these outcomes? a. A multiple of 3 or a multiple of 7, P(multiple of 3 or multiple of 7) b. P(even or odd) c. P(prime or 1) d. How did you find the probabilities of these events? Answer A: number of multiples of 3 or 7 from 1 to 20 total numbers = 20 p(multiples of 3 or 7)=(no of multiples of 3 or 7)/(total numbers) = 8/20 = 2/5 B: Number of even or odd numbers = 20 total numbers = 20 p(even or odd)=(number of even or odd)/(total numbers) = 20/20 = 1 C: no. of prime numbers and 1 total numbers = 20 p(prime or 1) = (number of prime numbers or 1)/(total numbers) = 9/20 D: probability could be found by general probability which says p(event)=(no.of times event occur)/(total no. of events) = 20/20 = 1 Question 2 Find the following probabilities for the standard normal variable z: (a) P(-1.96 ≤ z ≤ 1.96) (b) P(z > 1.96) (c) P(z < -1.96) Answer (a) P(-1.96 ≤ z ≤ 1.96) Determine the corresponding probability using the normal probability table in the appendix. P(z < 1.96) is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix. P(z < 1.96) = 0.9750 P(z < -1.96) is given in the row starting with -1.9 and in the column starting with 0.06 of the standard normal probability P(z < -1.96) = 0.0250 The probability between two boundaries is the difference between the probabilities to the left of both boundaries. P(-1.96 ≤ z ≤ 1.96) = P(z < 1.96) - P(z < -1.96) = 0.9750 - 0.0250 = 0.9500 (b) P(z > 1.96) Determine the corresponding probability using the normal probability table in the appendix. P(z > 1.96) is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix. P(z < 1.96) = 0.9750 Use the Complement rule: P(A') = 1 - P(A) P(z > 1.96) = 1 − P(z < 1.96) = 1 - 0.9750 = 0.0250 (c) P(z < -1.96) Determine the corresponding probability using the normal probability table in the appendix. P(z > 1.96) is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix. P(z < -1.96) = 0.0250 Result: (a) 0.9500, (b) 0.0250, (c) 0.0250. Question 3 Giving a test to a group of students, the grades and gender are summarized below A B C Total Male 7 16 19 42 Female 10 3 18 31 Total 17 19 37 73 If one student is chosen at random, Find the probability that the student got a C: Find the probability that the student was male AND got a "C": Find the probability that the student was male OR got an "A": If one student is chosen at random, find the probability that the student got a C GIVEN they are male: Answer Step 1 Determine the probability that student got a C using the formula P = (Number of students who got C) / (Total number of students) P = 37 / 73 = 0.506 Determine the probability that the student was male AND got a "C". P = 19 / 73 = 0.2603 Step 2 Determine the probability that the student was male. P(male) = 42 / 73 = 0.5753 Determine the probability that student got an "A". P(A) = 17 / 73 = 0.232 Determine the that the student was male OR got an "A" by substituting values in the formula P = P(male) + P(A) - P(male and A) P(male and A) = P(male) * P(A) = 0.5753 * 0.232 = 0.1334 P = 0.5753 + 0.232 - 0.1334 = 0.6739 Determine the probability that student got a C if they are male. P = 19 / 42 = 0.452 Question 4 When a fair die is rolled 10 times, what is the variation in the number of times a 6 appears? Answer The variance of the number of successes in n Bernoulli trials is npq where n is the number of tries, p is the success probability, and q is the fail probability. In this question, a success is when we get a 6 on the dice with probability p = 1/6. Hence q=1 - p = 5/6. Also, we throw the dice 10 times hence the Variance is 10 * (1/6) * (5/6) = 50/36 = 25/18 Question 5 Express the confidence interval using the indicated format. (Based on the percentages of red, orange, yellow, and blue MMs, the confidence intervals are calculated.) Express 0.0434 < p < 0.217 in the form p ± E Answer Given: 0.0434 < p < 0.217 p lies exactly in the middle of the confidence interval and thus is the sum of the boundaries of the confidence interval divided by 2: p = (0.0434 + 0.217) / 2 = 0.2604 / 2 = 0.1302 The margin of error is equal to the difference between the boundaries, divided by two, and is equal to half the width of the confidence interval: E = (0.217 - 0.0434) / 2 = 0.1736 / 2 = 0.0868 The confidence interval can then be rewritten as 0.1302 ± 0.0868