You really need to check some definitions.

nCr means the number of ways of choosing r objects from n objects, where order doesn't matter. How does the number of ways of choosing 2 cases from 8 help? (it doesn't)

I gave you the answer in the last line of my previous post O_o. And in the one before that told you that nCr and nPr aren't needed.

@monoxide I know you got the answer but he clearly doesn't understand why

OmgIRok

Taiwan2699 Posts

February 23 2009 01:20 GMT

#22

On February 23 2009 10:16 Monoxide wrote:

Show nested quote +

yes... soo 6! / 8!

OH wAIT YEAAA Okay so the success rates are all the positions where its 1XXXXXXX8 and since 1 & 8 are fixed you just find how many possible ways you can arrange 23456 which is 6! YES I SEE WHAT YOU DID THERE

OmgIRok

Taiwan2699 Posts

February 23 2009 01:24 GMT

#23

Okay the next question is P(all evens in the correct position) and I got 4!/8! <-- correct answer amirite?

Nytefish

United Kingdom4282 Posts

February 23 2009 01:26 GMT

#24

Yeah, these are the "easy" questions right?
The later ones will need you to use nCr and nPr?

OmgIRok

Taiwan2699 Posts

February 23 2009 01:44 GMT

#25

On February 23 2009 10:26 Nytefish wrote:
Yeah, these are the "easy" questions right?
The later ones will need you to use nCr and nPr?

Um... I think we are supposed to use ncr and npr on all of them as a shortcut to the counting principle but yeah whatever cuz i know npr and ncr formulas [not exactly sure what they are used for yet though, pretty sure used for what i mentioned earlier]

OmgIRok

Taiwan2699 Posts

February 23 2009 01:55 GMT

#26

well on to the next ones!!
I already did some of them, please check my answers? XD
3 people are chosen randomly out of 3 sophomores and 3 juniors

#1 P(0 sophmores)
3C0 / 6C3 = 1/20
2 P(1 sophmore)
3C1 / 6C3 = 3/20
#3 P(2 sophomore)
3 C 2 / 6C3 = 3/20
#4 P(3 sophomores)
3C3 / 6C3 = 1/20
#5 P(2 juniors)
Currently on #5

huameng

United States1133 Posts

February 23 2009 02:00 GMT

#27

The P(0 sophomores) + P(1) + P(2) + P(3) should add up to 1, so your answers to 1-4 don't make sense.

They don't add up because there are more than 3 ways to make a group with 1 or 2 sophomores. You also have to account for the number of different ways to pick the juniors.

OmgIRok

Taiwan2699 Posts

February 23 2009 02:07 GMT

#28

=O
okay I found out that P(1) and p(2) sophomores is actually supposed to be 9/20 chance to be picked, but where is that logic found?

6C1 gives me 6
6C2 gives me 15 15-6 = 9

Malongo

Chile3479 Posts

February 23 2009 02:12 GMT

#29

P(1)=3C1*3C2/6C3=9 because there are 3C1 ways to choose 1 sophomore and 3C2 ways to choose the remaining juniors. The same applies to P(2) and then P(0) + P(1) + P(2) + P(3)=1
#5 is easy if you use the same its P(2)=3C1*3C2/6C3=9

OmgIRok

Taiwan2699 Posts

February 23 2009 02:14 GMT

#30

Oh I see. thankyou for explaining!

Malongo

Chile3479 Posts

February 23 2009 02:15 GMT

#31

I note a lack of understanding in the multiplication principle when counting. The basic is that if an event A can occur in a different ways and an event B can occur in b different ways THEN the numbers of ways that the event A and B can occur is a*b.

OmgIRok

Taiwan2699 Posts

February 23 2009 02:16 GMT

#32

On February 23 2009 11:15 malongo wrote:
I note a lack of understanding in the multiplication principle when counting. The basic is that if an event A can occur in a different ways and an event B can occur in b different ways THEN the numbers of ways that the event A and B can occur is a*b.

haha yeah totally forgot about counting principle X_X eff me

Abydos1

United States832 Posts

February 23 2009 02:52 GMT

#33

On February 23 2009 09:00 OmgIRok wrote:
#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
I got 1/70 for this one. I seriously doubt it's correct though.

#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.

Does it specifically state those are the only ones in the correct positions? Because P(1 & 8 correct) can have other seasons in correct spots the way you worded it.

DeathByMonkeys

United States742 Posts

February 23 2009 04:11 GMT

#34

the first one is

8 nPr 5 or 8x7x6x5

the second one is

8 nPr 1 or 8! or 8x7x6x5x4x3x2x1

Each problem follows the same principles; you have a 1/8 chance of randomly placing the first dvd correctly, then a 1/7 chance of the next dvd and so on...

Malongo

Chile3479 Posts

February 23 2009 04:14 GMT

#35

I dont know if you know this problem but its a nice problem about "simple" probabilities:
The birthday problem:
In a room there are n people, what is the probability that at least 2 of them have a common birthday?
hint
+ Show Spoiler +

use the complement probability, that is find the P(no pair of them have the same birthday), then P(at least 2 of them have a common birthday)= 1-P(no pair of them have the same birthday)

solution
+ Show Spoiler +

since there are 365 posible birthdays then the total number of cases is 365!. The number of ways that no posible repetition of birthdays occur is 365*364*...*(365-n+1) because the first one have 365 posible birthdays, the second has 364 posible birthdays (because we want that he doesnt have the same birthday as the first), the third 363 and so on.
then P(no common birthday)= 365*364*...*(365-n+1)/365!=(365Cn)*n!/365! and the desired probablity is 1-(365Cn)*n!/365!.
why is this problem interesting? because its highly counterintuitive, for n=40 P>0.5 so for n=90 its almost sure that there are 2 with the same birthday.

JeeJee

Canada5652 Posts

February 23 2009 04:33 GMT

#36

@malongo
check your equations i think you typo'd something
and 23 is the magic number not 40 last i checked

Monoxide

Canada1190 Posts

February 23 2009 04:55 GMT

#37

ya the number is 23... and 23 gives a result of just over 50% actually

Malongo

Chile3479 Posts

February 23 2009 05:05 GMT

#38

On February 23 2009 13:33 JeeJee wrote:
@malongo
check your equations i think you typo'd something
and 23 is the magic number not 40 last i checked

?? please what is wrong? i rechecked but cant find it. And for the actual numebers i just put numbers that i was sure to fit, i dont remember them from memory i took probabilities about3 years ago dont ask that much

ahaha