d. The numerator comes from part c and the denominator from part b: =. 14

Transcription

1 Solution to assignmet 2 a Because order is important, we ll use P 8, 8(7)() b Order doesn t matter here, so we use C, 59, c From each group we choose 2: 8, , d The numerator comes from part c and the denominator from part b: 59,775 e We use the same denominator as in part d We can have all zinfandel, all merlot, or all cabernet, so all same) all z) all m) all c) ,775 2 a 5 b 2 c 8 d # with at least on Sony total # - # with no Sony e at least one Sony) 55 exactly one Sony) only Sony is receiver) only Sony is CD player) only Sony is deck)

2 a 8,7 2 all from day shift) 8, 8, 8,7 2 b all from same shift) c at least two shifts represented) all from same shift) d Let A day shift unrepresented, A 2 swing shift unrepresented, and A graveyard shift unrepresented Then we wish A A 2 A ) A ) day unrepresented) all from swing and graveyard) A ), A 2 ), A ) 5, A A 2 ) all from graveyard) A A ), A 2 A ) 2, A A 2 A ), So A A 2 A )

3 8 f selecting 2-75 watt bulbs) g all three are the same) h

4 ! 9! 9, (!) i To examine exactly one, a 75 watt bulb must be chosen first ( ways to accomplish this) To examine exactly two, we must choose another wattage first, then a 75 watt ( 9 ways) Following the pattern, for exactly three, 9 8 ways; for four, ; for five, examine at least bulbs) examine 5 or less) examine exactly or 2 or or or 5) [one) two) five)] [ ] j If the A s are distinguishable from one another, and similarly for the B s, C s and D s, then there are 2! Possible chain molecules Six of these are: A A 2 A B 2 C C D C 2 D D 2 B B, A A A 2 B 2 C C D C 2 D D 2 B B A 2 A A B 2 C C D C 2 D D 2 B B, A 2 A A B 2 C C D C 2 D D 2 B B A A A 2 B 2 C C D C 2 D D 2 B B, A A 2 A B 2 C C D C 2 D D 2 B B These (!) differ only with respect to ordering of the A s In general, groups of chain molecules can be created such that within each group only the ordering of the A s is different When the A subscripts are suppressed, each group of collapses into a single molecule (B s, C s and D s are still distinguishable) At this point there are 2!! molecules Now 2 chain molecules suppressing subscripts on the B s, C s and D s in turn gives ultimately k Think of the group of A s as a single entity, and similarly for the B s, C s, and D s Then there are! Ways to order these entities, and thus! Molecules in which the A s are contiguous, the B s, C s, and D s are also Thus, all together) 9 2 Seats: J&P in &2) 7 J&P next to each other) J&P in &2) J&P in 5&) 5 at least one H next to his W) no H next to his W)

5 We count the # of ways of no H next to his W as follows: # if orderings without a H-W pair in seats # and and no H next to his W * * 2 # 8 * pair, # can t put the mate of seat #2 here or else a H-W pair would be in #5 and # of orderings without a H-W pair in seats # and, and no H next to his W 2 # # can t be mate of person in seat # or #2 So, # of seating arrangements with no H next to W at least one H next to his W) - And no H next to his W), so 8 A A2 ) l A 2 A ) 5 A ) 2 m A A 2 A A ) 8 2 n We want P[(exactly one) (at least one)] at least one) A A 2 A ) Also notice that the intersection of the two events is just the st event, since exactly one is totally contained in at least one So P[(exactly one) (at least one)] 57 o The pieces of this equation can be found in your answers to exercise 2 (section 22): A A2 A ) 5 A A A2 ) 8 A A ) 2 5 p M LS PR) 5, directly from the table of probabilities q M Pr) M,Pr,LS) M,Pr,SS) 572 r SS) sum of 9 probabilities in SS table 5, LS) 5

6 s M) Pr) t M SS Pl) M SS Pl) SS Pl) SS M Pl) 8 u SS M Pl) M Pl) 8 LS M Pl) - SS M Pl) - 55