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    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?

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    Quote Originally Posted by leo View Post
    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?

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    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)

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

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    Quote Originally Posted by leo View Post
    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)

    ----------------

    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

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

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

    -----
    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

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    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.

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    Quote Originally Posted by leo View Post
    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

    add 1/210 and 15/210

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

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    Quote Originally Posted by leo View Post
    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

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

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    Quote Originally Posted by leo View Post
    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

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    regarding the second question, quoting your answer below, is the final answer 0.1429?

    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)

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    Quote Originally Posted by leo View Post
    regarding the second question, quoting your answer below, is the final answer 0.1429?

    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)

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