9. A local postal carrier distributes first-class letters, advertisements, and m
ID: 3196524 • Letter: 9
Question
9. A local postal carrier distributes first-class letters, advertisements, and magazines. For a certain day, she distributed the following numbers of each type of item. First-class Delivered to letters ds agazines Home Business 325 732 406 1021 203 97 If an item of mail is selected at random, find these probabilities. a. The item went to a home. b. The item was an ad, or it went to a business. c. The item was a first-class letter, or it went to a home. 10. Twenty workers are to be assigned to 20 different jobs, one to each job. How many different assignments are possible? 11. John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums?Explanation / Answer
Answer 9. Total number of mails (including all categories) = 325+406+203+732+1021+97 = 2784
(a) Total items(mails) that went to a home = 325+406+203 = 934
--> Probability that item(mail) went to a home = 934/2784 = 0.335
(b) Total items(mails) that were either an ad or went to a business = 406+1021+732+97 = 2256
--> Probability that item(mail) were either an ad or went to a business = 2256/2784 = 0.810
(c) Total items(mails) that were either first-class letters or went to home = 325+406+203+732 = 1666
--> Probability that items(mails) that were either first-class letters or went to home = 1666/2784 = 0.598
Answer 10. Let 20 workers be W1, W2, ... , W20 and 20 jobs be J1, J2, ... , J20. Now when assigning jobs to workers, W1 may be assigned any of J1 -- J20 (20 ways to assign), then W2 may be assigned any of remaining 19 jobs, W3 may be assigned any of remaining 18 jobs, ... W20 may be assigned the last remaining job.
So total number of unique assignments of jobs to workers possible = 20x19x18x...x3x2x1 = 20! (where " ! " stands for factorial) (20! = 2432902008176640000)
Answer 11. John may be assigned any 1 out of 4 instruments, then Jim can be assigned any 1 out of remaining 3 instruments, Jay may be assigned any 1 out of remaining 2 instruments, Jack would then be assigned remaining 1 instrument. So number of different arrangements possible = 4x3x2x1 = 4! (where " ! " stands for factorial) = 24
If jay and jack can play only piano or drums, then jay be assigned either a piano or drum (2 ways) and other instrument(piano or drum) may be assigned to jack. Total ways to assign instruments to jay and jack = 2! = 2
Now if it is necessary that each person gets exactly 1 instrument to play, then john and jim can take remaining 2 instruments in 2! = 2 ways, and total number of different arrangements = 2! x 2! = 4
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.