Question

Help! Help! Help!

4. Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time.
a) In how many ways can they arrive?
b) In how many ways can Jean arrive first?
c) Find the probability that Jean will arrive first.
d) In how many ways can Jean arrive first and Keith last?
e) Find the probability that Jean will arrive first and Keith will arrive last.

Answer 1

ohh boys i am studying the same thing these days... 1 thing is clear that this question is to be done by (not-replacement) method!!!

Answer 2

i think A is done like this but im not sure 6! = 720.

Answer 3

idk

Answer 4

A is certainly 6! = 720

Answer 5

@Sufiyancs what do you think is the answer in (B)

Answer 6

I think b would be 5! =120

Answer 7

man i am a bit weak at probability question and i dont want to misguide him :\

Answer 8

you are right again @Fletchnuts

Answer 9

The probability part is the only part that stumps me.

Answer 10

since (A) has all the ways they can arrive, then that is the denominator in the probability problem

Answer 11

can you help me step by step with C?

Answer 12

the numerator in (C) is the number of ways that jean will arrive first
the denominator is ALL the different ways these people can arrive (which is the answer in A)

Answer 13

so 120/720 or 1/6?

Answer 14

good.

Answer 15

what about D?

Answer 16

since jean is first and keith is last, your concern is the remaining four. in how many ways the four of them can arrive?

Answer 17

so you just do 4! ?

Answer 18

yeah

Answer 19

is this the numerator for E?

Answer 20

right again.

Answer 21

what the denominator?

Answer 22

whats*

Answer 23

ALL the possible ways ALL SIX OF THEM can arrive (result in A)

Answer 24

is it 720?

Answer 25

ok i got 1/30. thanks so much

Answer 26

YW