Question

Probability: 15 new students are to be distributed evenly among three classes. There are 3 girls among the new students.
a) What is the probability all 3 girls are assigned to the same class?
b) What is the probability that each class gets one girl?
(Work to follow)

Answer 1

For a, multinomial:
Number of ways students can be assigned without restriction:
\[\left[\begin{matrix} & 15 & \\ 3 & 3 & 3\end{matrix}\right]\] = 756,756

Answer 2

There are 3 classes to assign, so
\[3*\left(\begin{matrix}12 \\ 2\end{matrix}\right) \left(\begin{matrix}10 \\ 5\end{matrix}\right)\left(\begin{matrix}5 \\ 5\end{matrix}\right)\] = 49,896 ways to assign the three girls to one class room, therefore
P(all girls assigned to same class) = 49,896/156,156 = .0659

Answer 3

For b)
there are \[\left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}12 \\ 4\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right)\left(\begin{matrix}8 \\ 4\end{matrix}\right)\left(\begin{matrix}1 \\ 1\end{matrix}\right)\left(\begin{matrix}4 \\ 4 \end{matrix}\right)= 3*495*2*70*1*1= 207,900\] ways to distribute 1 girl to each class.
Thus P(one girl to each new class) = 207900/756,756 = .2747

Did I miss anything?

Answer 4

I wish I could tell if you did or not :) i take it this is past elementary statistics

Answer 5

It's the first week of probability. ;) Thanks anyway.

Answer 6

ok i got something different for first one but i didn't use multinomial. i did this

Answer 7

lets assume all girls get assigned to room 1 probability is
\[\frac{3}{15}\times \frac{2}{14}\times\frac{1}{13}\]

Answer 8

so multiply this by 3 because the events "all girls get assigned to room one" and "all girls get assigned to room 2" and "all girls assigned to room 3" are disjoint

Answer 9

so i get
\[\frac{18}{2730}=.00659\] rounded

Answer 10

which looks suspiciously like what you got but i think my decimal is correct and you are off by one place

Answer 11

Hmm. That is interesting....I will play around with it more. Did you agree with part b?

Answer 12

second one i think is right but i would do it differently. i think you just need to use your calculator again for first one. i bet we got same answer

Answer 13

I did...I realize I typed the denominator incorrectly (should be 756,756) but I still get .0659.

Answer 14

first of all \[\dbinom{1}{1}=\dbinom{4}{4}=1\] so forget the last terms

Answer 15

sell maybe your denominator is off.

Answer 16

Where did you get 18 in part a?

Answer 17

lets compute the probability that 1 girl gets assigned to room 1. it is
\[\frac{3}{15}\times \frac{12}{14}\times \frac{11}{12}\] and then the probability that one girl is assigned to room 2 given that one girl was assigned to room one is
\[\frac{2}{11}\times \frac{10}{10}\times \frac{9}{9}=\frac{2}{11}\]

Answer 18

oh i got 18 because i computed the probability that 3 girls got in room 1. that probability is
\[\frac{6}{2730}\] but there are three rooms so i multiplied by 3 to get
\[\frac{18}{2730}\]

Answer 19

i get to multiply by three because as i said the events are disjoint

Answer 20

ok

Answer 21

ok and finally if there is one girl in room 1 and one girl in room 2 there will be one girl in room 3 with probability 1. so my answer is
\[\frac{3}{15}\times \frac{12}{14}\times\frac{11}{13}\times \frac{2}{11}\]

Answer 22

let's see if it is the same as your answer. the problem with doing it with multinomial is that a huge compound fraction is forced on you so calculation is a bear. but if you do it without multiplying out you will see a plethora of cancellation and probably get exactly what i get

Answer 23

This is a frustrating thing in Probability (to me), there are many different ways to get the same answer. Arg! I don't think we agree but I'm not sure why. Thanks for your help, feel free to move on if you'd like. ;)

Answer 24

i get
\[\frac{12}{455}=.02637\]\]

Answer 25

oh crapola!

Answer 26

i am so sorry

Answer 27

i was putting 3 people in a class instead of 5!

Answer 28

lets knock the first one out again quickly.

Answer 29

Hahahaha awesome. Maybe that will help me out some cause I could not figure out what you were doing. Hehe.

Answer 30

ok 3 girls in same class. class 1 that is. you get
\[\frac{3}{15}\times \frac{2}{14}\times \frac{1}{13}\times 10\] damn you are right and i am wrong for sure sorry to waste your time, except i still think this is easier. then ten because there are ten ways to put 3 girls in 5 slots

Answer 31

mutiply this by 3 and get exactly your answer because my decimal was off not yours

Answer 32

it is not
\[\frac{18}{2730}\] but rather
\[\frac{180}{2730}=\frac{18}{273}=.0659\] what you said! so sorry

Answer 33

That is helpful. So you are getting your fractions by choosing girls instead of classrooms, basically, right?

Answer 34

yes. i put them all in one classroom and then multiply by 3

Answer 35

but like an idiot i forgot there were 5 people in the class!

Answer 36

so i had to multiply by ten, the number of ways to choose 3 from 5

Answer 37

Awesome. definitely quicker/easier than multinomial. I like having a different method to check. Right... 5C3. Makes sense. Thank you for sticking with it. These can take so damn long.

Answer 38

i am so sorry i forgot there were 5 people in a class and not 3. must have screwed you all up

Answer 39

ok now lets do last one quickly. we put 1 girl in class 1

Answer 40

Well, I wasn't convinced I was wrong since I couldn't talk myself into agreeing with your numbers so it wasn't too bad! ;)

Answer 41

\[\frac{3}{15}\times \frac{12}{14}\times \frac{11}{13}\times \frac{10}{12}\times \frac{9}{11}\times 5\]
and now we put 1 girl in class 2 given there is a girl in class 1
\[\frac{2}{10}\times \frac{8}{9}\times \frac{7}{8}\times \frac{6}{7}\times \frac{5}{6}\times 5\]

Answer 42

we do a ton of cancellation and then multiply these together.

Answer 43

aaaaand we agree, @ .2747. Sweet! Thanks bunches.

Answer 44

after cancelling everything in sight i get
\[\frac{5\times 5}{7\times 13}=\frac{25}{91}=.2747\]

Answer 45

sorry i didn't just look at what you wrote, pretended i got it and said "yes you are right"

Answer 46

ha! Thanks again for your help, I really appreciate it.

Answer 47

i still like this way though. hate dealing with huge numbers in numerator and denominator. you are welcome

Answer 48

Yes, it definitely simplifies some things. I always like having a new way to do something (especially when it saves time).