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Math Help - Probability Involving Permutations and Combinations

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    Probability Involving Permutations and Combinations

    Having a bit of trouble with this one. Can anyone help?

    Many thanks.

    Q. Assuming a newborn child is equally likely to be a girl or a boy, find the probability that, in a family of 5 children, there are more girls than boys.

    P(2 boys) = 1 - P(no boys) - P(1 boy) = 1 - (1/2 x 1/2 x 1/2 x 1/2 1/2) - 1/2 = 1 - 1/32 - 1/2 = 15/32

    Ans.: (From text book): 1/2
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    Re: Probability Involving Permutations and Combinations

    Quote Originally Posted by GrigOrig99 View Post
    Q. Assuming a newborn child is equally likely to be a girl or a boy, find the probability that, in a family of 5 children, there are more girls than boys.
    P(2 boys) = 1 - P(no boys) - P(1 boy) = 1 - (1/2 x 1/2 x 1/2 x 1/2 1/2) - 1/2 = 1 - 1/32 - 1/2 = 15/32
    Ans.: (From text book): 1/2
    The text book is correct.
    There can be zero, one, or two boys.
    \sum\limits_{k = 0}^2 {\binom{5}{k}{{\left( {0.5} \right)}^5}}  = \frac{1}{2}
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    Re: Probability Involving Permutations and Combinations

    Great, thank you.
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    Re: Probability Involving Permutations and Combinations

    Hi guys,
    I'm not a math student, in fact math is not a strong subject for me at all, quite the opposite. I'm a graduate computing science student and I could use some help with a problem I'm having. It's gonna be hard to explain given I'm not very mathematical so please bare with me. Ok so here's my problem.

    I have a program that asks a user a number of questions. For each question there are 3 possible answers, answers 1, 2, and 3. These answers have a value associated with them, a score if you will. Answer 1 has a value of 16, answer 2 has a value of 8, and answer 3 a value of 4. The layout would be something like this

    Question1: is this patent likely to be unique
    answer1: yes (value of 16)
    answer2: maybe (value of 8)
    answer3: no (value of 4)

    A user can only pick one answer per question, and remember there are 3 questions. So lets say the user picked answer 1 for all three questions, their score would be 48 (16x3).

    So what I need to do here is, calculate a result based on their score after they've answered the questions. So out of a possible highest score of 48, what percentage did the user get. If the user chose answer 1 for all three questions, the result is 100% obviously. If they chose answer1 for two of the questions, and answer3 for one of the questions, then the user scored 16 + 16 + 4 which is 36 out of a possible 48, 75%.....duh! I know this might seem pointless so far, stick with me as I need to explain this in order for my problem to make sense. So the result can mean different things depending on the percentage. There are 3 possible types of result, an SR1, SR2, and SR3. Don't worry what these mean, they make sense to the users. So if the user scored 75% or over the result is an SR1, if they scored 42% or over the result is an SR2, anything below that is an SR3. So it's like this

    SR1 >= 75%
    SR2 >= 42% and < 75%
    SR3 < 42%

    So if you're still with me, thanks. Here's where it gets a bit tricky for me. At the moment the program is asking the user 3 questions, but that number may increase and I need to allow for that in my score calculation algorithm. At the moment, with 3 questions the likelihood of an SR1 result, given all possible permutations of user selected answers, is roughly .33. The other probabilities are roughly,
    SR2 = .44
    SR3 = .22

    What I'd like to know is, how scalable is this with the current setup. I've noticed by scaling up from 3 questions to 8 questions the probabilities do change, as you'd expect I suppose. Is there anything I can do with my algorithm, any sort of formula to say, if the number of questions increases adjust the bands which SR1,2 or 3 can fall into accordingly so that the probabilities of any one of them occurring don't increase or decrease disproportionately with the number of questions being increased. Really not sure if any of that made sense. Any input at all is greatly appreciated.

    ps...would have started a new thread if I could find that button
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