Question

Page 578# 72
Every student in your history class is requite to present a project in front of the class . Each day , 4 students make their presentations in an order chosen at random by the teacher . You make your presentation on the first day.
a ) What is the probability that you are chosen to be the first or second presenter on the first day?
b) What is the probability that you are chosen to be second or third one the first day?. Compare your answer with that in part ( a )

Answer 1

Thanks

Answer 2

You're welcome :)

Answer 3

2 people have 2 different answer, I will post again to see what answer from difference people

Answer 4

14/24

Answer 5

whats ur question

Answer 6

the answer not match

Answer 7

I looked at the top and it lookes like kropot helped u

Answer 8

need explain and part b answer too

Answer 9

I did but don't have result

Answer 10

the formula is \[\frac{ n! }{ (n-r )!}\]

Answer 11

openstudy.com/updates/572ee741e4b06f843e1e4f15 XD i found ur question instead

Answer 12

I WILL BE BACK

Answer 13

some one answer 4! = 24 way
select first and second is 3x2 =6
second and third the same 3x2 = 6
than 12 out of 12/24 = 1/2 ?

Answer 14

I hat to do probability

Answer 15

7/12 and 1/2 which one ?

Answer 16

I don't understand dropout's reasoning.

Answer 17

I don't know either , I try post more to see another answer

Answer 18

Here are all the permutations of 4 numbers. Pretend your name is 1
count how many have 1 in the first or 2nd position
and divide by the total number
>> perms([1 2 3 4])
ans =

1 2 3 4
2 1 3 4
1 3 2 4
2 3 1 4
3 1 2 4
3 2 1 4
1 2 4 3
2 1 4 3
1 3 4 2
2 3 4 1
3 1 4 2
3 2 4 1
1 4 2 3
2 4 1 3
1 4 3 2
2 4 3 1
3 4 1 2
3 4 2 1
4 1 2 3
4 2 1 3
4 1 3 2
4 2 3 1
4 3 1 2
4 3 2 1

Answer 19

4 students = 4! = 24 ways
than find probability first + probability second
I'm not sure how to find probability os first and second

Answer 20

the brute force way is to count the number of permutations that has 1 in the first or second column. there are 12 of them
12/24 = ½

Answer 21

first or second is mean first + second

Answer 22

you said first is 1/2 ?

Answer 23

second the same 1/2 ?

Answer 24

6 have a 1 in the first column
6 have a 1 in the 2nd column

so 12 have a 1 in the first or 2nd column (meaning picked to be first or second)

Answer 25

a)
b)
the same answer 1/2?

Answer 26

Here are the 12 permutations
1 2 3 4
2 1 3 4
1 3 2 4
3 1 2 4
1 2 4 3
2 1 4 3
1 3 4 2
3 1 4 2
1 4 2 3
1 4 3 2
4 1 2 3
4 1 3 2

with 1 in the first or 2nd column

Answer 27

yes 2nd or 3rd is also `½

Answer 28

ok, I will let you know after my home work correct from teacher , can you help me more?

Answer 29

what is the question ?

Answer 30

let me attach file

Answer 31

I need help number 74, 75

Answer 32

74)
a= 3 C 2 ?
b+ 3P2 ?

Answer 33

74 a) would be 3 choose 3
(there are 3 colors, and we choose 3 of them
3C3 is 1
in other words, if we ignore order (we do in a combination), we get a red, green, blue

Answer 34

\[a) \frac{ 3! }{(3-2)!\times2! }\]

Answer 35

for 74 b), there are 3! permutations
3 ways to pick the first color, 2 for the 2nd and 1 for the 3rd:
3*2*1 or 3!

Answer 36

why do you think a is 3C2 ?

Answer 37

b) \[\frac{ 3! }{ (3-2)! }\]

Answer 38

a) the question ask how many combination than is C

Answer 39

b) the question ask how many permutations than is P

Answer 40

yes, but in a) it is 3C3 not 3C2

Answer 41

b) is 3P3 which is 3!/0! = 3! = 6

Answer 42

or for b) you can count how many there are by looking at the figure. It shows 6 permutations

Answer 43

FIRST draw is 3 color , SECOND DRAW IS 6 , each color out come 2
3 C2 ?

Answer 44

the bag has 3 marbles.
you choose 3 of them
so it is 3 choose 3 or 3C3

Answer 45

thanks , can you hep 75?

Answer 46

for 75 a) do you know the total number of permutations ?

Answer 47

\[\frac{ 10! }{(10-5)! }?\]

Answer 48

10! permutations total
(we don't care about intermission or how many are before the intermission)

next, we put your rival and you in position 4, 5
that leaves 8 people to go in the other positions, or 8! ways

the answer is 8!/10! = 1/(10*9) = 1/90

Answer 49

how you get 8?

Answer 50

you are told you and your rival are in fixed positions.
we have to count all the ways the other 8 people are arranged. (that would be 8!)

Answer 51

you are the last and you are rival performs immediately before you is mean , last and before is 2 ? than 10 -2 = 8?

Answer 52

yes

Answer 53

thanks ,
b) you are not the first than 10-1=9
9!/10! ?

Answer 54

9!/10! is the number of ways for you to be first. (we fix you in position 1, and find 9! ways to arrange the other 9 people)

to get the number where you are *not* first, is do 1 - 9!/10!

Answer 55

you should get
1 -1/10 = 0.9 for b)

Answer 56

don't understand part b

Answer 57

how you subtract 1 ?

Answer 58

if there are 10 people, and we put you first
that leaves 9 people that we can arrange.
so 9! ways for you to be first. ok with that?

Answer 59

the question as you are not first , why you said you first ?

Answer 60

the probability for being first is easy to figure out
and we know
Pr(being first) + Pr(not first) = 1
(one or the other must happen)
so
Pr(not first) = 1 - Pr(first)

Answer 61

if you want to figure out Pr(not first) we could do it this way:

out of 10 people, 9 can be in the first position (we have to leave you out)
there are now 9 people and 9 positions left, and 9! ways to fill them
so
9 * 9!
divide by 10! (number of ways to arrange 10 people)
we get
\[ \frac{9 \cdot 9!}{10!} = \frac{9}{10} = 0.9 \]

Answer 62

you are great, are you Math teacher ?

Answer 63

no

Answer 64

Can you tutor me sometime I will Pay ?

Answer 65

you don't have to pay. but I will help when I am around.

Answer 66

no, I love to pay I not compotable ask help but don't pay, I can pay a little bet to OK,can I email ?

Answer 67

I will send you a check