Solved Final Exam - Probability for Life Sciences Students | MATH 3C, Exams of Mathematics

Download Solved Final Exam - Probability for Life Sciences Students | MATH 3C and more Exams Mathematics in PDF only on Docsity! MATH 3C - Lecture 1 - Summer 2009 Practice Final Solutions - July 22, 2009 This is a closed-book and closed-note examination. graphing calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 8 problems for a total of 200 points. 1 2 1. (30 points) A worker has asked her supervisor for a letter of recom- mendation for a new job. She estimates that there is an 80% chance that she will get the job if she receives a strong recommendation, a 40% chance if she receives a moderately good recommendation, and a 20% chance if she receives a weak recommendation. She further estimates that the probabili- ties that the recommendation will be strong, moderate or weak are 0.7, 0.2, and 0.1 respectively. (a) What is the probability that she will receive the new job offer? (b) If she received the job offer, what is the probability that the recommen- dation was moderate? (c) If she did not receive the job offer, what is the probability that the rec- ommendation was strong? Solution. Let X be the indicator random variable on the type of recom- mendation the worker will receive, i.e. X = 1, 2, or 3 if the worker re- ceives good, moderate, or poor rec letter, respectively. We were given that P (X = 1) = 0.7, P (X = 2) = 0.2, and P (X = 3) = 0.1. Let Y be the indicator random variable for the job she gets. i.e. Y = 1, if she gets the job and Y = 0 otherwise. So we were further given that P (Y = 1|X = 1) = 0.8, P (Y = 1|X = 2) = 0.4, and P (Y = 1|X = 3) = 0.2. (a) P (Y = 1) = 3∑ i=1 P (Y = 1 ∩X = i) = 3∑ i=1 P (Y = 1|X = i)P (X = i). This result follows from the law of total probability. Thus, P (Y = 1) = (0.7)(0.8) + (0.2)(0.4) + (0.1)(0.2) = 0.66. (b) P (X = 2|Y = 1) = P (X = 2 ∩ Y = 1) P (Y = 1) = (0.2)(0.4) 0.66 = 4 33 ≈ 0.121 (c) P (X = 1|Y = 0) = P (Y = 0 ∩X = 1) P (Y = 0) = (0.7)(0.2) 0.34 = 7 17 ≈ 0.412 ¤ 5 4. (20 points) Suppose 3 balls are chosen without replacement from an urn consisting of 5 white and 3 red balls. Let X equal 1 if the first ball selected is white and 0 otherwise. Also let Y equal 1 if two out of three balls selected are red and zero otherwise. Find the joint probability mass function of X and Y . Solution. Note that the sample space is Ω = {WWW,WWR, WRW,RWW,RWR, RRW,RRR}. Since X and Y can only take on two values, 0 or 1, only three values need to be computed: P (X = 0, Y = 0), P (X = 1, Y = 0), P (X = 0, Y = 1), P (X = 1, Y = 1). P (X = 0, Y = 0): This is the occurrence that neither the conditions for X nor the conditions for Y occur. Taking a glance at our sample space, this only occurs with the draw RWW. And since we are drawing without replacement, P (RWW ) = 38 · 57 · 46 = 528 . P (X = 1, Y = 0): The draws that give this output are WWW,WRW , and WWR. Hence, P (X = 1, Y = 0) = P (WWW )+P (WRW )+P (WWR) = 5 8 ·4 7 ·3 6 + 5 8 ·3 7 ·4 6 + 5 8 ·4 7 ·3 6 = 5 28 P (X = 0, Y = 1): The draws that give this output are RRW,RRR, and RWR. P (X = 1, Y = 0) = P (RRW )+P (RRR)+P (RWR) = 3 8 ·2 7 ·5 6 + 3 8 ·2 7 ·1 6 + 3 8 ·5 7 ·2 6 = 11 56 P (X = 1, Y = 1): This only occurs with the draw WRR. P (WRR) = 5 8 · 37 · 26 = 556 . ¤ 6 5. (20 points) You arrive at a bus stop at 10 o’clock, knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30. (a) What is the probability that you will have to wait longer then 10 min- utes? (b) If at 10:15 the buss has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes? Solution. (a) If we graph the probability density function (using minutes as our units), we see that it’s 0 outside the interval [0, 30], and inside the interval, it is constantly 130 . If we want to know the probability that you will have to wait longer than 10 minutes, then we’re looking for P (X > 10), which is the area of a rectangle with length 20 (the distance from 10 to 30), and base 130 , which is equal to 2 3 . (b) We’re looking for P (X > 25|X > 15) = P (X > 25 ∩X > 15) P (X > 15) = P (X > 25) P (X > 15) = 5 30 / 1 2 = 1 3 . ¤ 7 6. (30 points) Suppose that the travel time from your home to your office is normally distributed with mean 40 minutes and standard deviation 7 minutes. (a) If you leave home at 8 A.M., what is the probability you get to the office by 8:45 A.M. (b) Find the probability that the travel time is between 35 and 45 minutes. (c) If you want to be 95% certain that you will not be late for an office appointment at 1 P.M., what is the latest time that you should leave home? Solution. Let X be normally distributed with mean 40 and standard de- viation 7, and let Z be standard normally distributed (mean 0, standard deviation 1). (a) P (X < 45) = P (Z < 45−407 ) = P (Z < .714) = F (.714) ≈ .7612, where F is the standard normal distribution F (x) = P (X < x). (b) P (35 < X < 45) = P (35−407 < Z < 45−40 7 ) = P (−.714 < Z < .714) = 2F (.714)− 1 ≈ .5224. (c) We’re looking for a number so that the probability of the travel time being less than that number is 95%. That is, a number k such that P (X < k) = .95. But P (X < k) = P (Z < k−407 ). According to the table for the standard normal distribution, P (Z < 1.65) ≈ .95, so k−407 = 1.65, or k = 51.55. This means that 95% of the time, the travel time is less than 51.55 minutes. Hence, if you leave by 12:09 P.M., the probability of arriving on time is 95%. ¤