Question

A company has 22 staff members - the owner, 3 senior executives, 14 junior executives and 4 support staff. A committee of 8 members of the staff must be convened and the members are selected at random.
a) what is the probability that there are fewer support staff on the committee than senior executives.
b) what is the probability that the owner in on the committee and there is at least one member from each of the 3 staff categories on the committee too?

Answer 1

First thing we need to know is how many committees of 8 people can be formed from randomly selecting 8 from 22.

Answer 2

22C8

Answer 3

agrreed

Answer 4

and that would be the n(s) for P(A) =n(A)/n(s) ; A = the event specified in question

Answer 5

Now, if there are fewer support staff than senior executives this can occur in a number of different ways. 0 support and 1,2 or 3 senior. 1 support and 2 or 3 senior. Or 2 support and 3 senior. This gives n(A)=6.

Answer 6

So, if my reasoning is correct, you have P(A)=6/22C8

Answer 7

no...the answer is 81549/319770

Answer 8

hmmm.

Answer 9

22C8 = 319770 so i dont know where 81549 comes from

Answer 10

oh...oooops yeah

Answer 11

is the question trying to say that with 3 execs, there is the possibility of no support staff or there will be 1 or 2?

Answer 12

so i think in a case of where there is 3 execs, then no. of ways is
3C3.15C5 + 3C3.4C1.15C4+3C3.4C2.15C3?

Answer 13

with 3 seniors AND zero support. We should have 15C5. With 3 seniors and 1 support: 16C4. With 3 seniors and 2 support, 17C3. So,
15C5 + 16C4 + 17C3=3003+1820+680=5503. We still have other possibilities to add to this. We could have 2 seniors with zero support. 2 seniors with 1 support. 1 senior with zero support. add them up...

Answer 14

ah yes...i get the logic now. How would you do the second part of the question?

Answer 15

hmmmm I still thing I missing something though after reading your last post.

Answer 16

with 3 seniors and zero support 3C3*15C5=15C5 is correct.

Answer 17

with three seniors and 1 support we would have 3C3*4C1*16C4

Answer 18

I got the answer! :D

Answer 19

sweet...am I on the right track at least? Im rusty on perms/combs/probability

Answer 20

yip :) thanks to your logic, thats how i got it

Answer 21

not sure how to do the second part...what would your logic be?

Answer 22

because the owner must the in the committee, we can only choose from 21 staff

Answer 23

the number of committees we can form with the owner on it is 21C7

Answer 24

and from the 7 staff chosen, we allocate their positions in the committee?

Answer 25

from the seven committee spots left over we need at LEAST one person from each category of employee. there are lots of possibilities ehre

Answer 26

so like 7C1.6C1.5.1 and then multiply 5C4?

Answer 27

for the case where we have one from each category

Answer 28

the amount of 68438 is what we are looking for in n(A)

Answer 29

well, we pretty much have to itemize each way of doing this

Answer 30

your teacher is cruel btw

Answer 31

i very much agree! and he sometimes never completes hard examples with us!

Answer 32

if we have to itemise everything...this will take so long and i have one more question to ask you (for the day)...should we just give up on that?

Answer 33

so, one way is 1 owner, 1 senior, 5 junior and 1 support. I get, 1C1*3C1*14C5*4C1

Answer 34

There must be a shortcut to this one...way too many possibilities

Answer 35

1 - the probability of the opposite?

Answer 36

yeah, that's what I was thinking...but...not seeing it

Answer 37

ok...we need at least 1 member from each group. So the opposite would be having zero members from each group

Answer 38

agreed

Answer 39

n(A) = 21C7?

Answer 40

wait, if none of them is chosen...only the owner would be in the group

Answer 41

so..starting with the seniors group. The probability of forming a committee with zero seniors 19C5. zero seniors and 1 owner: 18C4. At LEAST one senior 1-19C5.

Answer 42

do u mean 1 - 19C5 - other combinations too?

Answer 43

Now, the juniors. zero juniors and 1 owner, 7C8=not possible. We MUST have at least one junior on the committee. Thus, probability of having at least one junior =1

Answer 44

There's an error on my reasoning about the seniors group. The probability of having one owner and at least one senior is 1-P(one owner and zero seniors)=1-(18C7/22C8)

Answer 45

but we can't have a probability of 1...

Answer 46

So...the probability of having 1 owner and at least one junior and at least one senior is (1-(18C7/22C8)*1

Answer 47

now...last group. The prob of having 1 owner and at least one support is 1-P(one owner and zero support)=1-(17C7/22C8)

Answer 48

then you multiple them all together. overall probability is (1-(18C7/22C8))*(1-(17C7/22C8))

Answer 49

I'll let you compute this and see if it is correct (if not I think I know where my error likely will be)

Answer 50

and is not correct. the answer is suppose to be 0.2140

Answer 51

ok

Answer 52

ur answer gives 0.8457...

Answer 53

damn....tough one. I think your stuck itemizing them all and adding up the individual possibilities

Answer 54

wish I could be more help but I have to go to work

Answer 55

good luck :)

Answer 56

ok thanks. will u be online tomorrow?

Answer 57

yeah probably

Answer 58

i need to review this stuff a bit :(

Answer 59

is it ok if i ask you for help tomorrow?

Answer 60

yeah of course

Answer 61

No one likes perms and combs grrrrrrr

Answer 62

yep...I hated this stuff when I did it years ago and never looked back

Answer 63

but u seem really good with this anyways :)

Answer 64

i ll go study other stufff now. ciao ciao :)