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'''for e, i thought of it as there are 4 spots available for one of the letters a or z. spots 1,2,25,26 since there has to be at least 23 letters inbetween. so you have 4 choose 1. then the other letters has 2 spaces to choose from that are at least 23 letters apart from the other letter. then counted the number of permutations the other 24 letters could have. as for f... i'm stuck on that too.''' | '''for e, i thought of it as there are 4 spots available for one of the letters a or z. spots 1,2,25,26 since there has to be at least 23 letters inbetween. so you have 4 choose 1. then the other letters has 2 spaces to choose from that are at least 23 letters apart from the other letter. then counted the number of permutations the other 24 letters could have. as for f... i'm stuck on that too.''' | ||
| + | -I think for e you only have 2 spots for the 25 letter "unit" (a-----z or z-----a) either it starts at the 1 spot or two spot. | ||
for f, actually there's a simple way to look at this, the probability of it being before ab, in between ab and after ab are all equally likely. and ba is already considered in this case. So there are only 3 cases and all are equally likely. | for f, actually there's a simple way to look at this, the probability of it being before ab, in between ab and after ab are all equally likely. and ba is already considered in this case. So there are only 3 cases and all are equally likely. | ||
| + | -No, because a and b don't have to be together. | ||
[[Category:MA375Spring2010Walther]] | [[Category:MA375Spring2010Walther]] | ||
Revision as of 17:51, 24 February 2010
HW6MA375S10
6.2 - 2, 10, 12, 18, 24, 26, 30
Section 6.2
2.
10. I know d is 1 - answer to c. Any help on c?
for c, think of a & z are one letter "az". but then you also have to consider the reverse "za". then count how many permutations you can do with that.
Any ideas on part e and/or f?
for e, i thought of it as there are 4 spots available for one of the letters a or z. spots 1,2,25,26 since there has to be at least 23 letters inbetween. so you have 4 choose 1. then the other letters has 2 spaces to choose from that are at least 23 letters apart from the other letter. then counted the number of permutations the other 24 letters could have. as for f... i'm stuck on that too. -I think for e you only have 2 spots for the 25 letter "unit" (a-----z or z-----a) either it starts at the 1 spot or two spot.
for f, actually there's a simple way to look at this, the probability of it being before ab, in between ab and after ab are all equally likely. and ba is already considered in this case. So there are only 3 cases and all are equally likely. -No, because a and b don't have to be together.
12.
18. b & c are confusing me. (7 choose n) (1/7)^n, and 7^n are used, but in what form?
If you look at the part of example 13 on pg 410 where they use the pn's this may help you Example 13 helps a lot!! I didn't understand how to do it till i read it. Thanks!!
24. Did anyone else get the answer to be (1/2^5)/(1/2) = 1/16?
yes, by using P(E|F)= P(E intersection F)/P(F)
26.
30. Any hints for part b)? (b) is very similar to (a)
part c) is it just 1/2*1/2^2*1/2^3.....*1/2^10? cause that answer is 2.775E-17 and I don't believe that is a correct solution.
it is a really low number, but it does make sense if you look at the problem
