A Truck and a VanDate: 07/11/99 at 04:02:44 From: Pekkle Subject: Probability A small cartage company has two vehicles, a truck and a van. The truck is used 75% of the time; both vehicles are used together 40% of the time; and neither of the vehicles is used for 10% of the time. Let T be the event "the truck is used" and V be the event "the van is used." a) complete the statement P(T U V)= P(T) + _____? Hence find the proportion of the time that the van is used. b) Show that the two vehicles are not statistically independent in their use. c) Given that the van is being used, find the probability that the truck is also being used. (Use a probability tree.) Thanks. Date: 07/11/99 at 07:24:50 From: Doctor Anthony Subject: Re: Probability >Let T be the event "the truck is used" and V be the event "the van >is used." P(T or V) = P(T) + P(V) - P(T and V) 0.9 = 0.75 + P(V) - 0.4 P(V) = 0.9 - 0.75 + 0.4 = 0.55 So the van is used 55% of the time. >b) Show that the two vehicles are not statistically independent in >their use. To be statistically independent we require P(T and V) = P(T) x P(V) 0.4 = 0.75 x 0.55 = 0.4125 NOT TRUE so the two are NOT statistically independent. >c) Given that the van is being used, find the probability that the >truck is also being used. (Use a probability tree.) | T not T | Total ------|----------------|----------- V |0.4 |0.55 not V | 0.10 |0.45 ------|----------------|------------ Total |0.75 0.25 |1.00 The above table can be filled in as shown. It is now possible to complete it to make the row and column totals agree. We complete the table as follows: | T not T | Total ------|----------------|----------- V |0.4 0.15 |0.55 not V |0.35 0.10 |0.45 ------|----------------|------------ Total |0.75 0.25 |1.00 And now from the table we can answer the last question. "Given that the van is being used" means we are in row V with a total sample space of 0.55. Then the probability that the truck is also being used is 0.4/0.55 = 0.7272.. = 8/11 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 07/13/99 at 04:39:29 From: Pekkle Subject: Probability Thank you for answering my question. I have another: The following table shows how many of the 1992 graduates from a university had jobs by May. (The following tables are together.) Occupation Agriculture Business Education Arts Science Engineering ---------------------------------------------------------------------- Permanent 209 110 158 17 48 85 employment ---------------------------------------------------------------------- Seeking 47 17 21 37 62 11 employment ---------------------------------------------------------------------- Not seeking 22 15 8 15 14 5 employment ---------------------------------------------------------------------- Totals 278 142 187 69 124 101 ---------------------------------------------------------------------- Occupation Social Sciences Medicine Totals ------------------------------------------------------- Permanent 96 39 762 employment ------------------------------------------------------- Seeking 68 31 267 employment ------------------------------------------------------- Not seeking 31 7 117 employment ------------------------------------------------------- Totals 195 50 1146 Consider the following events: A: that a randomly selected graduate is either from the education or social sciences faculties B: that a randomly selected graduate is seeking employment (required to show your reasoning and working clearly for parts (a) and (b). a) The events A and B are: A. Complementary but not independent B. Mutually exclusive but not independent C. Mutually exclusive and independent D. Neither mutually exclusive nor independent E. Not mutually exclusive, but independent b) The conditional probability of A happening, given that B has happened, and the probability of either A or B or both happening are: A. P(A/B)=0.333 P(A or B)=0.566 B. P(A/B)=0.333 P(A or B)=0.489 C. P(A/B)=0.078 P(A or B)=0.566 D. P(A/B)=0.078 P(A or B)=0.489 E. P(A/B)=0.333 P(A or B)=0.078 Thanks a lot. Date: 07/13/99 at 16:52:21 From: Doctor Anthony Subject: Re: Probability >Consider the following events: >A: that a randomly selected graduate is either from the education or >social sciences faculties = 187 + 195 = 382 out of total of 1146 >B: that a randomly selected graduate is seeking employment = 267 out of 1146 >(required to show your reasoning and working clearly for parts (a) >and (b). > >a) The events A and B are: > A. Complementary but not independent To be complementary you require P(B) = 1 - P(A) That is 267/1146 = 1 - 382/1146 This is clearly NOT TRUE To be independent you require P(A and B) = P(A) x P(B) Then P(A and B) = (21 + 68)/1146 = 89/1146 and P(A) x P(B) = 382/1146 x 267/1146 = 1/3 x 89/382 = 89/1146 since these two probabilities are equal, the two events are independent. We conclude that the events are NOT complementary but ARE independent. > B. Mutually exclusive but not independent Since there are numbers that are in BOTH A and B the events are NOT mutually exclusive. We have already seen that the events are independent, so wrong again on both counts. > C. Mutually exclusive and independent Not Mutually exclusive - see previous answer. Yes, they are independent. > D. Neither mutually exclusive nor independent Not mutually exclusive but they ARE independent. > E. Not mutually exclusive, but independent Correct. Not mutually exclusive, and they are independent. > >b) The conditional probability of A happening, given that B has >happened, and the probability of either A or B or both happening are: If B has happened the sample space is limited to the 'Seeking Employment' row. The numbers in this row for Education or Soc.Science = 21 + 68 = 89 So P(A|B) = 89/267 = 1/3 P(A or B) = P(A) + P(B) - P(A and B) = 382/1146 + 267/1146 - 89/1146 = 560/1146 = 0.488656 So the two answers are P(A|B) = 1/3 and P(A or B) = 0.488656 These agree with the answers given under (B) below. > B. P(A/B)=0.333 P(A or B)=0.489 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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