Results 1 to 4 of 4

Math Help - Probability problem

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    19

    Probability problem

    I would appreciate if you solve this too. Thanks.


    Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37

    Cool

    Quote Originally Posted by chil2e View Post
    Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?
    Let's first determine the probability that the first person picks his own name and the remaining 4 people do NOT pick their own names.

    Knowing this, we cast the first person aside for right now, and consider the the following question: Four people remove their name tags and put them in a hat. Then they each randomly choose one of the tags. What is the probability that exactly none picks their own name?

    Let E denote the event that NO matches occur. Since the probability depends on n, the number of people choosing, we write  P_n=P(E) . We are going to go about this solution recursively, so we begin by conditioning on whether or not the first (of 4) people draws his own name tag. Let's denote these events I \text{  and  } I^c . So,

    <br />
P_n = P(E) = P(E|I) P(I) + P(E|I^c) P(I^c)<br />

    Note that  P(E|I) = 0 , and thus
     P_n = P(E|I^c) P(I^c) = P(E|I^c) (\tfrac{n-1}{n}) .

    Now,  P(E|I^c) is the probability that none of n-1 people pick their own name tag when selecting from a hat containing n-1 tags with exactly one of the tags not belonging to any of the n-1 people. (So, there 1 "extra person" whose name tag is not in the hat, and there is 1 extra tag in the hat which does not belong to any of the men choosing.) This can happen in two mutually exclusive ways:

    1) No matches and extra man does NOT select the extra hat (hat of man who chose first). The probability of this event is P_{n-1} (which we get if we assume the extra hat belongs the extra person).

    2) No matches and extra man DOES INDEED select the extra hat. The probability of this event is \tfrac{1}{n-1}P_{n-2} .

    Since the two ways described above are mutually exclusive,  P(E|I^c) is simply the sum of the two probabilities; i.e.,
     P(E|I^c) = P_{n-1} + \tfrac{1}{n-1} P_{n-2}
    .

    Now, we can plug this information into the equation for  P_n :

    <br />
P_n = P(E|I^c) \tfrac{n-1}{n} = \tfrac{n-1}{n} P_{n-1} + \tfrac{1}{n} P_{n-2} \implies P_n - P{n-1} = -\tfrac{1}{n} (P_{n-1} - P_{n-2})<br />

    Aha! Recursion! It can do the rest of our dirty work.

    So we have  P_n is the probability of n people all selecting a name tag that is NOT their own (no matches). Thus,

    <br />
P_1 = 0<br />

    <br />
P_2 = \tfrac{1}{2}<br />

    Now we take advantage of that recursion.

    <br />
P_3 - P_2 = -\tfrac{p_2 - P_1}{3} = -\tfrac{1}{3!} \implies P_3 = \tfrac{1}{2!} - \tfrac{1}{3!}<br />

    <br />
P_4 - P_3 = -\tfrac{p_3 - P_2}{4} = -\tfrac{1}{4!} \implies P_4 = \tfrac{1}{2!} - \tfrac{1}{3!} + \tfrac{1}{4!} <br />

    Therefore, the probability of 4 people each selecting name tags (each person's name included) that are NOT their own is  P_4 = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}  = 0.375 .

    Now, let's bring back the real first person to select from the hat. The probability that he, selecting first, selected his name own tag is 1/5. So the probability of (HIT, MISS, MISS, MISS, MISS) is (3/8)/(1/5). However, not only must we consider this person being "the one" to select his own hat (while the others don't), we must also consider the separate events whereby each of the other four people is "the only one" to select his name tag. So, we multiply (3/8)/(1/5) by (1/5), which cancels out the (1/5) in (3/8)/(1/5). Thereforeforeforefore, the probability of EXACTLY ONE person selecting his own name tag in our problem is 3/8 = 0.375.

    -Andy
    Last edited by mr fantastic; January 15th 2010 at 03:53 PM. Reason: Fixed latex
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,614
    Thanks
    1581
    Awards
    1
    That reply is too long to even quote.
    There arer five ways to select the one getting his/her own tag.
    There are D(4)=9 derangements the other four.
    So the probability of exactly one is \frac{5\cdot 9}{5!}=0.375.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Plato View Post
    There are D(4)=9 derangements the other four.
    That was neat!

    Here is another logical way...

    Label them A,B,C,D,E.
    If A strikes, then the other 4 must miss.
    The ways this can happen are as follows....

    A picks A in all following cases
    B picks C, C picks B so D must pick E to leave E picking D
    B picks C, C picks D so D must pick E to leave E picking B
    B picks C, C picks E so D must pick B to leave E picking D.

    In the second of these cases, if D picked B, E would be left picking E.

    There are 3 ways, for which the probabilities are...

    \frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}

    in all cases.

    Repeat the process for B picks D and B picks E,
    therefore, we treble the probabilty for the B picks C case.
    Finally multiply by 5 as any of the 5 can strike.

    P=5(3)(3){\frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}}=\frac{3}{8}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 0
    Last Post: October 8th 2009, 08:45 AM
  3. Probability problem 1
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 9th 2009, 11:38 PM
  4. Probability problem 3
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 9th 2009, 11:38 PM
  5. Probability problem 4
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 9th 2009, 11:38 PM

Search Tags


/mathhelpforum @mathhelpforum