Question

zarkon , I need help Please!

Answer 1

Maybe you should just ask your question and hope someone can answer.

Answer 2

Janice has 8 DVD cases on a shelf, one for each season of her favorite
TV show. Her brother accidentally knocks them off the shelf onto the floor.
When her brother puts them back on the shelf, he does not pay attention to
the season numbers and puts the cases back on the shelf randomly. Find
each probability.
1. P(all even-numbered seasons followed by all odd-numbered seasons)
2. P(all even-numbered seasons in the correct position)
3. P(seasons 5 through 8 in any order followed by seasons 1 through 4 in
any order)

Answer 3

maybe if i write it with words it will be more clear, because i guess it didn't work the first time. sometimes words escape me. lets do the first one slow. all even numbers followed by all odd numbers.
first one is even, 4 out of 8 or 4/8
second one is even given that first one is even is 3/7 since there are thee even and 7 to choose from
third one is even given first two are is 2/6 and finally fourth one is even is 1/5

now you are done, because if all the even ones are first necessarily all odds will be second. so your answer is
\[\frac{4}{8}\times \frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 4

1)?
the key is: 1/8!

Answer 5

really??

Answer 6

funny because i saw that answer elsewhere and assumed it was wrong. there are 8! ways to arrange these dvds yes?

Answer 7

so to say the answer is
\[\frac{1}{8!}\] means there is only one way to do this. but that is silly

Answer 8

I don't need number 2, how about 3 ) ?

Answer 9

3) should be the same.

Answer 10

1) and 3) the same?

Answer 11

just replace the word "even" by 5 through 8 and the word "odd" by 1 through 4

Answer 12

there are 4 even and there are 4 labeled 5 though 8 etc. these should be the same. i don't understand why it would say the answer to the first one is 1/8!

Answer 13

maybe i am reading the question wrong

Answer 14

nope i am sticking with my answer. there is clearly more than one way to do this yes?

Answer 15

I'm not good at probability

Answer 16

where is the answer key? is it in the book? on line?

Answer 17

I disagree Satellite. There is only one way to do this.

Answer 18

ooh? then i am wrong

Answer 19

You must pick all the even seasons first, and all the odd seasons second.

Answer 20

but the way i read it i will give you two
2, 4, 6, 8
4, 2, 6, 8

Answer 21

2) 4!/8!
3) (4!*4!)/8!

Answer 22

3) is the same answer i wrote

Answer 23

From the set of 4 even numbered we will chose 4 disks. There is only 1 way.
Then from the set of 4 odd numbered we will chose 4 disks. There is only 1 way for this also.

We don't care what order they are in inside.

Answer 24

For the purposes of this problem, choosing 2,4,6,8 is the same as choosing 4,3,6,8

Answer 25

\[\frac{4!\times 4!}{8!}=\frac{4!}{8\times 7\times 6\times 5}\]
\[=\frac{4}{8}\times \frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\] they are the same

Answer 26

That being said. I am also not great at probability.

Answer 27

ok so we agree that the answer to c) is what you wrote and what i wrote, because they are the same yes?

Answer 28

I think I get now: denominator is 8!
5 through 8= 5,6,7,8,= 4!
1 through 4=1,2,3,4 = 4! than
(4!*4!)/8!

Answer 29

Hrm. But I'm also feeling like C should have the same answer as A. So I'm probably totally wrong ;p

Answer 30

@polpak i can demonstrate more than one way to do number 1 and in fact number 1 and number 3 are the same with the words changed from "even" to 1-4 etc

Answer 31

so i am sticking with my answer.

Answer 32

you can think of it as
\[\frac{4!\times 4!}{8!}\] if you like but i think it is more intuitively clear if i write
\[\frac{4}{8}\times \frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 33

no number two is different

Answer 34

1) and 3) = 4!/8!
or 3) = (4!*4!)/8!

Answer 35

I don't need 2)

Answer 36

number two requires the even numbered ones to be in the correct position, not any positiion

Answer 37

2) and 3) the same ?
4!/8!

Answer 38

number two is
\[\frac{4!}{8!}=\frac{1}{8}\times \frac{1}{7}\times\frac{1}{6}\times \frac{1}{5}\]

Answer 39

no one and 3 are the same

Answer 40

any order. just separated

Answer 41

how 3 )?

Answer 42

one and three are
\[\frac{4!\times 4!}{8!}\]

Answer 43

1. P(all even-numbered seasons followed by all odd-numbered seasons)
3. P(seasons 5 through 8 in any order followed by seasons 1 through 4 in any order)

the way i read these they are identical

Answer 44

there is no order given for the first one
the second one say "any order" explicitly. so just replace the word "even-numbered" by the words "season 5 through 8" and they are the same

Answer 45

thank, I'm very bad at probability, I better with geometry.

Answer 46

good luck

Answer 47

here is a site with the answer you had for the first one, but i am sure it is wrong.
http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.473393.html
it says "# of ways to pick 4 even: 1" but that is not correct. you can pick
2, 4, 6 8
4 ,2 6, 8,
2, 4, 8, 6
etc. there are 4! ways to do it, not one

Answer 48

#1) has to be

\[\frac{4!\cdot4!}{8!}\]

Answer 49

bless you.

Answer 50

;)

Answer 51

i was thinking very simply. don't know if you read my post. i explained it in words

Answer 52

maybe if i write it with words it will be more clear, because i guess it didn't work the first time. sometimes words escape me. lets do the first one slow. all even numbers followed by all odd numbers. first one is even, 4 out of 8 or 4/8 second one is even given that first one is even is 3/7 since there are thee even and 7 to choose from third one is even given first two are is 2/6 and finally fourth one is even is 1/5 now you are done, because if all the even ones are first necessarily all odds will be second. so your answer is

Answer 53

if they wanted all even (in order) then all odd (in order ) then we would have

\[\frac{1}{8!}\]

Answer 54

\[\frac{4}{8}\times \frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 55

yep

Answer 56

3)?

Answer 57

i thought i was losing my mind because this looked fairly straight forward but "answer key" said 1/8! which is silly so did web post i found

Answer 58

same as #1

Answer 59

bless you again. just change the words

Answer 60

1/8! is clearly wrong

Answer 61

2)
4!/8! ?

Answer 62

right. that would mean there is only one way to do it. do not trust everything you read on the web. especially here!

Answer 63

yes

Answer 64

bless you again. but i would write it as
\[\frac{1}{8}\times \frac{1}{7}\times \frac{1}{6}\times \frac{1}{5}\]

Answer 65

by reasoning it out, rather than some formula

Answer 66

in fact though it is clearly correct i think it rather silly to write
\[\frac{4!}{8!}\]

Answer 67

I hate do probability, I rather take geometry instate

Answer 68

how about

\[\frac{1}{8}\cdot\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}\]

I prefer \cdot ;)

Answer 69

ooh nifty

Answer 70

cdot has one few letter than times too!

Answer 71

you can also have

\[\cdots\]

\cdots

Answer 72

we need latex forum
tricks and tips
maybe a wolram one too

Answer 73

\[\ddots\]

\[\vdots\]

Answer 74

now you are showing off.

give...

Answer 75

lol

Answer 76

Josh types the five entries in the bibliography of his term paper in random order , forgetting that thet should be in alphabetical oder by author ,What is the probability that he actually typed them in a alphabetical order?

Answer 77

1/5!

Answer 78

only one way to type in alphabetical order, 5! ways to arrange them

Answer 79

\[\checkmark\]

Answer 80

\[\check\]

Answer 81

no i guess not

Answer 82

\checkmark

Answer 83

sorry I post wrong one, this one
The game of euchre (YOO ker) is played using only the 9s, 10s,
jacks, queens, kings, and aces from a standard deck of cards. Find the
probability of being dealt a 5-card hand containing all four suits.

Answer 84

\[\checkmark\]

Answer 85

\[\vdots\]

Answer 86

\[\clubsuit\diamondsuit\heartsuit\spadesuit\]

Answer 87

wow how you post the suite?

Answer 88

\clubsuit\diamondsuit\heartsuit\spadesuit

Answer 89

i got \[\clubsuit\diamondsuit\heartsuit\spadesuit\]

Answer 90

ok i am off let me know if you get the same answer i got. i think it was
\[\frac{6^4\cdot 5\cdot4\cdot3\cdot2}{24\times 23\times 22\times 21\times 20}\]

Answer 91

hmm..I'm not getting that

Answer 92

I may skip this question, thank all your guy

Answer 93

how about this

Describe an event that has probability of 0 and an event that has probability of 1.

probability made me headache

Answer 94

roll a 6-sided die...what is the probability you get a 0

flip a coin...what is the probability you will get a head or tail

Answer 95

roll=1/6
flip a coin=1/2

Answer 96

P(get a 0)=0
P(H or T)=1

Answer 97

are you still around , I will post some more probability later , I run some errant and be back later.
Thanks

Answer 98

here is my answer to your previous problem...