Revision as of 16:57, 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.

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