HW6MA375S10
6.2 - 2, 10, 12, 18, 24, 26, 30
Do we have to post for a grade this week?
I also I'd like to know,
When is the last week for rhea project for a grade?
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.
Here are my thoughts; can someone tell me if they seem right or wrong? Since the positions of a, b, and z are totally random, z has an equal chance (1/2) of coming before or after a. And, z has an equal chance (1/2) of coming before or after b. If the events "z comes before a" and "z comes before b" are independent, then the probability of the event "z comes before a AND z comes before b" should be 1/2 * 1/2 = 1/4. The problem is, I'm not so sure they're independent, so I don't know that you can just multiply the probabilities together and get the answer...
But there are 26! different outcomes, how many outcomes involve z before a and b?
There are 26 choose 3 ways to get the three, then 23! ways to order them, and z is either before or after a and b. Since there are 26! total ways you divide by that. So... (26 choose 3)*(23!)*(2)/26! is what I got and that simplifies to about 1/3. (I went to his office hours the other day and that is what he told me.)
Hmm... in fact, it doesn't just simplify to *about* 1/3; it simplifies to *exactly* 1/3, which seems to agree with the answer given by the first person who posted about part f. But I still don't understand the reasoning they gave. How would the following three events be equally likely? - "z comes before both a and b" - "z comes between a and b" - "z comes after both a and b"
Think of it as the different ways to order a, b, and z: abz, baz, azb, bza, zab, zba and the probabilities are obvious.
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)
Is parts b) like part a) where instead of 1/2^10 the the ratio of a single toss is no longer 1/2 but 3/5?
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.
Yea, that is the correct way to do it.
I did it the same, there isn't really any other way to do it, it's a lot like a and b
it is a really low number, but it does make sense if you look at the problem
