Question

Michel, Jonathan, and two of their friends (say it was Jimmy and Matthew) each ride their bikes to school. If they have an equally-likely chance of arriving first, what is the probability that Jonathan will arrive first an Michel will arrive second?

Answer 1

2) Glen received 6 birthday cards. If he is equally likely to read the cards in any order, what is the probability he reads the card from his parents and card from his sister before the other cards?
3) Codes: For exercises 4-6, use the following information. A bank gives each new customer a 4-digit code number which allows the new customer to create their own password. The code number is assigned randomly from the digits 1,3,5, and 7, and no digit is repeated.
4) What is the probability that the code number for a new customer will begin or end with a 7?
5) What is the probability that the code number will NOT contain a 5?
6) What is the probability that the ocde number will start with a 371?
Thanks for your time and effort...:DDDD

Answer 2

bumping in 1 seconds :DDD help is on da way:DDD

Answer 3

oh my goodness.. um... just a sec then.

Answer 4

..OMFGOMFGOMFG.... r u a gurl?????

Answer 5

nope, male

Answer 6

For the first one, the chance being equally likely between all of them, would be 1/4 for each
Then, once one person arrives first, I believe there is still an equal chance between the remaining people to make second. So, you'd have three people left once the first person makes it. So, a 1/3 chance with three people.

Answer 7

oh oops sorry.. oh gosh...*embarrasment*

Answer 8

Then, its the probability of both events happening. That should be multiplication of the events' probabilities, I do believe.

Answer 9

mmmhmmm ur rite :DDD
nerd B)

Answer 10

What are you working on tonite?

Answer 11

...old hmwk..dat i don't get...D:

Answer 12

same situation with the next problem
it is equally likely to pull out any of the 6 cards, so uniform distribution is 1/6 probability for each card

if we first pull one card and then without replacing it, pull another, its 1/5 (because its one person's card in particular, and a sample space of 5)
1/6 * 1/5

Answer 13

So we think the chances that we predict the order of the first two is about 1 in 12?

Answer 14

Yes, concur with AccessDenied.

Answer 15

so da second prolems answer is 1/30???

Answer 16

I would think so.

Answer 17

Problem 4: How many different outcomes are there for the first digit?

Answer 18

...4?

Answer 19

Yep. Any of them more likely than the others?

Answer 20

i don't think so..

Answer 21

is it3/ 4?

Answer 22

brb

Answer 23

Correct again! So, that indicates a one of four chance that the first digit is seven. Similarly, there is a one in four chance that the last digit is seven. In fact, the seven will be somewhere, and since there are only four places to go, and two of them are the first or last spots, there is a one in two chance that seven will either begin or end the code.

Answer 24

oh thanks so den for number 5..it's sorta easy rite?
is da answer like, um, like,uh, like 0/4?

Answer 25

OK, question five: There are four positions in the code, and four numbers to go into it. None are repeated. Can we ever leave one of the digits (in the question, five) out?

Answer 26

yeah, sounds about right to me

Answer 27

By George, I think she has it!!

Answer 28

Question six: How many choices for the first digit?

Answer 29

one fourth.. i fink...
hu's george?

Answer 30

Actually four choices...How many remain for the second digit?

Answer 31

By George must be an expression older than you are....

Answer 32

oh cuz den my best friend nerdy angelica has a crush on dis guy named geroge in our class and den i h8 him bcuz he rejected her at dance.. so pissed at him da't y i h8 george names... D:

Answer 33

Back to math....

Answer 34

Take another tack.... if the code starts with 371, what is the whole code?

Answer 35

lool okay? :D

Answer 36

duh 3715.. duh ..duh.duh

Answer 37

isn't it obvious?

Answer 38

OK so there is only one way to have the first three digits 371; how many ways can the numbers be arranged?

Answer 39

one fourth

Answer 40

...i fink...

Answer 41

How about four (ways to pick the first number) times three (ways to pick the second number) times two (ways to pick the third number) times one (the last number is the one left) or twenty four. So there is one of twenty four chance the code starts with 371.

Answer 42

oh oooppppsss it's late.. thank yoyu soooooiooooooooooooooooooooooooooo vvvveeeeeeerrrrryyyyyyyyy mmmuuuuhhhhhhccccccccccccchhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! <333333333333333333333333333333333

Answer 43

i fink u get it :]

Answer 44

You're killing me....LOL

Answer 45

Past my bedtime; good nite.

Answer 46

i had NO idea, teachers could say LOL
lol :P
and no offense tho

Answer 47

Goodnight. :)
I will now return to other things.

Answer 48

yes good nite @accessdenied, and @AnimalAin <333

Answer 49

@ animalain, could you explain problem 4 again? thanks

Answer 50

I'm too old to be up this late....

Answer 51

Think of it this way.... We established that each number is used in every code, right?

Answer 52

mhuh

Answer 53

So the seven will certainly go to one of the four spots, right?

Answer 54

yeppos :]

Answer 55

Well, half of the spots are middle spots (second or third) and half are end spots (first or last). That means there is a one out of two chance that the seven will be in the first or last spot. Get it?

Answer 56

say i'm stupiud..i AM
uh.. could you explain like how you woud to a ...3rd grader????

Answer 57

I mean, it's going somewhere, right? Half of the spots meet the specification (in the first or last spot) so the probability is one of two.

Answer 58

maybe it's because i am stupiud.. or maybe i'm just trying to fall asleep by accident while my grampa gives me pectures.. oh man

Answer 59

lectures*
but oh well he let me do it 2morro
bye???o.e >.<

Answer 60

zzzzzz
-.-

Answer 61

Recall that this is the question:

What is the probability that the code number for a new customer will begin or end with a 7?

We know that the seven will go in the code somewhere (that was problem five). There are four places it can go. Two of them meet the specification of the problem (first or last spots) and two don't (second or third spots). So two out of four possibilities, all with the same likelyhood, mean the chance of "success" is two of four, or one of two.

Answer 62

the probability that Jonathan will arrive first an Michel will arrive second is 1/6

Answer 63

thanks, @worldmath

Answer 64

and thanks a lot to YOU @animalain!
<333
and sorry i wuz like very sleepy so ya i just fell over... :P

Answer 65

so, @animalain, is it 3/4, 1/4, or 1/2?.
..1/2?

Answer 66

Answers in recap (you could look up there, you know...)

1) 1/6
2) 1/30
3) No question, no answer.
4) 1/2
5) 0
6) 1/24

Answer 67

Hmm.. how is the probability of that first one 1/6?

Answer 68

Sorry, sloppy documentation. It was yesterday, after all....

1) 1/12

Answer 69

actually, worldmath also mentioned 1/6

Answer 70

oh sorry never mind my grampa just told me...thanks guys..for the explanations :D
and sorry i wasn't ...paying attention..and feel asleep yesterday, while you were explaining stuff to me..oh well...
2morro's skool anyway..
sigh...

Answer 71

It's ok. I think we've all faded off at one time or another...ZZZZZ....