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Thread: Contactlens prescription value probability

  1. #1
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    Contactlens prescription value probability

    Math question here, hope someone can help me out.
    If a contactlens company has 213696 different prescription combinations possible. And the company sells a total of 3 billion contacts a year. What are the chances in both % and "1:x", that a specific prescription combination is used by someone? (Leaving out factors such as specific value ranges someone is likely to have a prescription for.)

    Hope someone can help me out, thanks in advance!
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    Re: Contactlens prescription value probability

    The number of times a specific prescription is selected out of the 3 billion is a binomial distribution with parameters

    $p = \dfrac{1}{213696}$
    $n = 3\times 10^9$

    $P[\text{specific prescription is used at least once}] = \\

    1 - P[\text{prescription is used by no one}] = \\

    1 - (1-p)^n = \\

    1 =100\% = 1:1 \\

    \text{It's not truly 1 but it's so close you couldn't fit an atom between it and 1}$
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  3. #3
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    Re: Contactlens prescription value probability

    Thanks so much sir, for your explaination! The calculations aren't fully comprehensible for me, but I'll try to do some exploring myself.
    Best regards.
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  4. #4
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    Re: Contactlens prescription value probability

    Quote Originally Posted by turbopriest View Post
    Thanks so much sir, for your explaination! The calculations aren't fully comprehensible for me, but I'll try to do some exploring myself.
    Best regards.
    Let's look at the answer term by term

    Quote Originally Posted by romsek View Post
    The number of times a specific prescription is selected out of the 3 billion is a binomial distribution with parameters
    I'll leave you to wiki what a binomial distribution is. Basically if you have an event that can succeed with some probability $p$ and you run $n$ trials of that event,
    the number of successes in those n trials has a binomial distribution.

    $p = \dfrac{1}{213696}$
    $n = 3\times 10^9$
    These two lines just compute the $p$ and $n$ mentioned above. The probability you select one prescription from the lot of the is just $\dfrac{1}{\text{total number of prescriptions}}$
    $n$ is the 3 billion sold you mention.

    $P[\text{specific prescription is used at least once}] = $
    This is what the problem asks for

    $1 - P[\text{prescription is used by no one}] = $
    The probability that an event occurs is 1 minus the probability that it doesn't occur. That's just basic probability.

    $1 - (1-p)^n = $
    The probability that our chosen prescription is not selected in 3 billion tries is just the probability that it's not selected, i.e. $(1-p)$, multiplied times itself 3 billion times.
    That's the probability that the chosen prescription was not selected in 3 billion tries. One minus this is the probability that it was. I.e. what the problem asks for.

    $1 =100\% = 1:1 \\
    \text{It's not truly 1 but it's so close you couldn't fit an atom between it and 1}$
    Evaluating this it was 0.99999 (hundreds of 9s, maybe thousands, I didn't count)
    which is close enough to 1 to call it 1.
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