STAT/MATH 9 A - Autumn Quarter 015 - Midterm Name: Student ID Number: Problem 1 5 Total Points Directions. Complete all questions. You may use a scientific calculator during this examination; graphing calculators and other electronic devices are not allowed and should be turned off for the duration of the exam. If you use trial-and-error, a guess-and-check method, or numerical approximation except when asked for when an exact method is available, you will not receive full credit. When you are asked to compute probabilities, provide the exact numerical solution, unless otherwise stated. Show and explain all work and write in complete sentences to receive credit. clearly the events you are considering. Particularly, specify You have 50 minutes to complete the exam. 1
Problem 1. 5 pts Let E, F, G and H be four mutually independent events, with PE PF PG 1/ and PH /. a Compute PE F H 1 pt Answer. Since events are mutually independent, we have : PE F H PEPF PH 1 1 7 0.07 b What is the probability that at least one of the events occur? pts Answer. It is easier to compute the probability of the complement of the requested event, that is that none of the events occur. Thus, the desired probability is : 1 PE c F c G c H c 1 PE c PF c PG c PH c 1 1 1 1 1 1 1 1 7 81 0.901 c What is the probability that exactly three of the four events occur? pts Answer. The event A that exactly three of the four events occur can be written as the union of mutually exclusive events as follows : A E c F G H E F c G H E F G c H E F G H c. Hence, we have : PA PE c F G H + PE F c G H + PE F G c H +PE F G H c PE c PF PGPH + PEPF c PGPH + PEPF PG c PH +PEPF PGPH c 1 1 1 1 + 1 + 1 1 1 1 1 81 0.16 1 1 1 + 1 1 1 1 Problem. 5 pts 5% of the population have video game console X, 15% have console Y and 5% own both consoles. According to a study related to video game addiction, 10% of possessors of console X exhibit symptoms of addiction, this rate is 0% for owners of console Y and 0% for possessors of both consoles. If a randomly selected person owns at least one of the two consoles, what is the probability that this person is an addict? Answer. Let us denote the following events: A: the person is an addict X: the person owns console X Y : the person owns console Y We want to find PA X Y :
PA X Y PA X Y PX Y PA X A Y PX + PY PX Y PA X + PA Y PA X Y PX + PY PX Y PA XPX + PA Y PY PA X Y PX Y PX + PY PX Y 0.1 0.5 + 0. 0.15 0. 0.05 0.5 + 0.15 0.05 1 9 0.111 Problem. 5 pts In a city, there are three tennis clubs A, B and C that are composed of 55, 8 and 6 players, respectively. A player can only be member of one tennis club., 5 and 7 players of clubs A, B and C respectively are sponsored. One player from each club is selected to play a tournament. What is the probability that the player chosen from club B is sponsored given that exactly two selected players are sponsored? Answer. Let us define the following events : S A : The player selected from club A is sponsored S B : The player selected from club B is sponsored S C : The player selected from club C is sponsored The event that exactly two selected players are sponsored, denoted S, can be expressed as the union of three mutually exclusive events : S S A S B S c C S A S c B S C S c A S B S C. Therefore, the desired probability is : PS B S PS B S PS PS A S B SC c + PSc A S B S C PS A S B SC c + PS A SB c S C + PSA c S B S C The fact that a player can only be a member of one tennis club implies that the selection from each tennis club is independent from each other. Therefore the probabilities appearing in the fraction above can easily be computed as follows : PS A S B S c C PS A PS B PS c C 55 5 8 56 6 1 198 PS A S c B S C PS A PS c BPS C 55 8 7 6 790 PS c A S B S C PS c APS B PS C 5 55 5 8 7 6 1 1188 Plugging in the numbers and simplifying gives the final solution : PS B S 80/509 0.77
Problem. 7 pts On a game show you play the following game. First you flip a fair coin. Based on the result of that flip you play another game. If the coin comes up heads then you roll ten dice. In this case you win if at least two of the ten dice are sixes. If the coin comes up tails then you pick two cards from a standard deck of fifty-two cards. In this case you win if both of the cards are of the same suit. You do not need to give numerical solutions for this problem. a When you roll ten dice, what is the probability that exactly k dice are sixes, with 0 k 10? pts Answer. For a given k 0... 10, let E k be the event that exactly k of the ten dice land on six. In that case, the remaining 10 k dice do not show six. The probability that a die lands on 6 is 1/6, the probability that a die does not land on 6 is thus 1 1/6. Therefore, by independence of the 10 throws, we have that : PE k 10 k 1 6 k 1 1 6 10 k 10 k 5 10 k 6 10 b If you pick two cards from a standard deck of fifty-two cards, what is the probability that both of the cards are of the same suit? pts Answer. Let S be the event that both cards drawn from a standard deck of fifty-two cards are of the same suit. For a fixed suit for instance, hearts, there are 1 ways to select two cards from that given suit. Since there are four suits, the desired probability is : PS 1 5 Note that we obtain the same result if we compute the probability of drawing a first card of a given suit and that the second card is of the same suit. PS 1 5 1 51 c What is the probability of winning the game described above? pts Answer. Let us denote W the event that one wins the game. The outcomes of the coin s toss are denoted H and T for heads and tails, respectively. According to the law of total probability, we have : PW PW {H} + PW {T } PW {H}P{H} + PW {T }P{T } When the coin comes up heads, the event of winning occurs if at least two of the ten dice are sixes, 10 which is the union of the following mutually exclusive events : k E k, where E k is defined in question a. When the coin comes up tails, the event of winning occurs if both of the drawn cards are of the same suit, which is exactly event S defined in question b. Hence, we have : PW PW {H}P{H} + PW {T }P{T } 1 1 10 k { 10 k PE k + 1 PS 6 10 + 1 } 5 10 k 5 10 k
Problem 5. 8 pts The 10 students of a school have to take one language class and can choose among three languages, namely French, Japanese and Spanish. 0 students took a French class, chose Japanese and 57 Spanish. At the end of the year, 9, 19 and 8 students of the French, Japanese and Spanish classes respectively passed their exams. a If we know that a randomly selected student passed his or her exam, what is the probability that he or she took the French class? pts Answer. Let F, J and S be the events that a student studies French, Japanese and Spanish, respectively and let us denote P the event that the student passes his or her exam. We are given the following information : PP F 9/10, PP J 19/10 and PP S 8/10 /5. We want to find PF P. This is a direct application of the definition of conditional probability: PF P PF P PP PF P PP F + PP J + PP S 9/10 9/96 0.0 9/10 + 19/10 + /5 b The 1 members of a student board are randomly selected from the 10 students of the school. What is the probability that the board is composed of French learners, Spanish learners and Japanese learners? pts Answer. The set of all possible boards consists of all 10 1 possible combinations of 1 people selected from the 10 students of the school. Let E be the event that consists of selecting French learners, Spanish learners and Japanese learners, this can be done in 0 57 ways. Hence, we have PE 0 c Alain took the French class. Is the event that Alain is a member of the board independent of the event that the board has exactly French learners? Justify your answer. pts 10 1 Answer. Let us denote A the event that Alain is a member of the board and let B be the event that the board has exactly French learners. We need to determine whether the identity PA B PAPB holds. If Alain is a member of the board, there only remains to pick 11 people from the 119 remaining people to complete the board. Therefore, 1 119 PA 1 11 1 10 10 1 Event B occurs when exactly French learners are chosen among 0 and 8 non-french learners men are selected among 80. 80 8 PB 10 1 A B is the event that Alain is a member of the board and that the board has exactly French learners, meaning that other French learners are to be selected among 9 and 8 non-french learners among 80 to complete the board. PA B 0 57 1 9 80 1 8 10 1 By noticing that 1 10 0 9, one can see that the identity PA B PAPB holds, entailing that the event that Alain is a member of the board is thus independent of the event that the board has exactly French learners. 5