1. Probability Problem

Pls help...........

My problem exercise goes like this - An employee of the records office at a certain school currently has 10 forms on his desk awaiting processing, six of these are withdrawal petitions and the other four are course substitution request.

a. if he randomly select six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk?

b. suppose he has time to process only four of these forms before leaving for the day. If four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?

2. Originally Posted by leo
Pls help...........

My problem exercise goes like this - An employee of the records office at a certain school currently has 10 forms on his desk awaiting processing, six of these are withdrawal petitions and the other four are course substitution request.

a. if he randomly select six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk?

b. suppose he has time to process only four of these forms before leaving for the day. If four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?
What seems to be the problem...Where are you stuck?

3. You have to think about what's being asked. There's no formula we can recommend for this, but a good approach is to break it up into parts.

a. P[only one of the two types remains on desk] = P[only withdrawals remain on desk] + P[only substitutions remain on desk]
To find P[only withdrawals remain on desk], think about it: to have only withdrawals remaining, you need to pick all 4 of the substitutions, and 2 of the withdrawals.

b. P[alternating for 4 forms] = P[W, S, W, S] + P[S, W, S, W]
where W=withdrawal, S=submission
We have P[W, S, W, S] = (6/10) * (4/9) * (5/8) * (3/7)

4. thanks joe... I am stuck on letter a. question, don't know where to start

5. Originally Posted by leo
thanks joe... I am stuck on letter a. question, don't know where to start
for P(pick all 6 W's)

how many ways can you choose the 6 from the 10 total

when picking 6 how many ways can you pick all 6 W's

(remember order doesn't matter here)

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for P(pick all 4 S's)

again how many ways can you choose the 6 from the 10 total

how many ways to pick the 4 S's and how many ways to pick 2 W's out of the 6

6. can you show me the formula please so I can start..thanks

7. Originally Posted by leo
can you show me the formula please so I can start..thanks
First off: This problem uses the hypergeometric distribution.

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so if we let nCr=n!/[r!(n-r)!]

so nCr is the number of ways to choose r objects from n when order doesn't matter

so we need to compute 10C6,6C6,4C4, and 6C2 and then put them together correctly

8. I'm still confused. What I only know is the probability of getting 6 withdrawal forms is 0.00476... can you expound your computation please.

9. Originally Posted by leo
I'm still confused. What I only know is the probability of getting 6 withdrawal forms is 0.00476... can you expound your computation please.

P(6W when 6 are pulled)=6C6/10C6=1/210=.00476 which you got

P(4S when 6 puled)=P(4S and 2W)=4C4*6C2/10C6=1*15/210=15/210

10. i got the same answer 0.00476 for 4C4*6C2/10C6. don't know how you got 15/210

11. Originally Posted by leo
i got the same answer 0.00476 for 4C4*6C2/10C6. don't know how you got 15/210

4C4=1

6C2=6*5/2=3*5=15

10C6=10*9*8*7/4*3*2*1=10*3*7=210

12. so the probability that only one of the two types of froms remain on the desk is 0.0762. is that right?

13. Originally Posted by leo
so the probability that only one of the two types of froms remain on the desk is 0.0762. is that right?
yes..8/105~.0762

b. P[alternating for 4 forms] = P[W, S, W, S] + P[S, W, S, W]
where W=withdrawal, S=submission
We have P[W, S, W, S] = (6/10) * (4/9) * (5/8) * (3/7)

15. Originally Posted by leo

b. P[alternating for 4 forms] = P[W, S, W, S] + P[S, W, S, W]
where W=withdrawal, S=submission
We have P[W, S, W, S] = (6/10) * (4/9) * (5/8) * (3/7)
the answer to b is .1429 (1/7 to be exact)