Re: Probability Question



Gregory L. Hansen wrote:
> In article <1121180529.889735.150720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> Schoenfeld <schoenfeld1@xxxxxxxxx> wrote:
> >
> >
> >Gregory L. Hansen wrote:
> >> In article <1121158112.199275.147680@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> >> Schoenfeld <schoenfeld1@xxxxxxxxx> wrote:
> >> >
> >> >
> >> >Gregory L. Hansen wrote:
> >> >> I've been doing some probability problems, and there's one I just can't
> >> >> figure out. It doesn't seem like it should be hard.
> >> >>
> >> >> "A hatcheck girl loses track and passes out the four hats she has at
> >> >> random. What is the probability that none of the four owners gets the
> >> >> right hat?"
> >> >>
> >> >> The answer is supposed to be 3/8, but even when I tried a brute-force
> >> >> approach of writing out each of 24 permutations I found 10/24.
> >> >>
> >> >> Any hints?
> >> >
> >> >Trivial problem.
> >> >
> >> >By an owner drawing a wrong hat then the following owners have higher
> >> >chance of drawing a wrong hat.
> >> >
> >> >Probability of person 1 getting wrong hat is P(1) = 1 - 1/4 = 3/4.
> >> >
> >> >Probability of person 2 _also_ getting wrong hat is P(2) = P(1)
> >> >(1-1/(3-1)) = 3/8.
> >> >
> >> >Obviously at this point if two people have already drawn wrong hats,
> >> >then the remaining two owners are guaranteed to draw wrongly thus the
> >> >answer is 3/8.
> >>
> >> Unless the order drawn is 2134.
> >
> >No. The order is defined by the hat drawing (i.e. person 1 is 1st
> >person to draw hat,etc).
>
> Hat belonging to person 2,
> hat belonging to person 1,
> hat belonging to person 3,
> hat belonging to person 4.

Please read what I said properly, it is the correct solution.


> --
> "The hardest conviction to get into the mind of the beginner is that the
> education he is receiving in college is not a medical course but a life
> course for which the work of a few years under teachers is but a
> preparation." -- Sir William Osler

.

Relevant Pages

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