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Thread: Probability Problem

  1. #1
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    Probability Problem

    Not an expert in Probability, I am stuck on ii and iii question of this problem. Help will be greatly appreciated

    An MBA professor will be teaching a Finance elective course next semester, and

    broadcasts the course description to 50 students. Given the mathematical demands of the

    course, there is only a 5% chance that a student will sign up for the elective.

    i. What is the probability that 5 or more students from this set of 50 students will sign

    up?

    ii. How many students from this set of 50 can he expect to sign up for the Finance

    elective?

    iii. The professor needs at least 10 students in order for the elective not to be canceled.

    What is the smallest number of students to whom he needs to send the course

    description, in order to have a 95% probability of the course not being canceled?
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  2. #2
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    Re: Probability Problem

    This is a binomial distribution problem.

    The number of students that sign up for the elective, $k$, is a random variable that has a binomial distribution,

    $Binomial(50, 0.05)$, i.e.

    $P[k \text{ students sign up for elective}] = \binom{50}{k}(0.05)^k (0.95)^{n-k}$

    1) $P[k\geq 5] = 1 - P[k \leq 4] = 1 - \displaystyle{\sum_{k=0}^4}~P[k]$

    2) $E[K] = \displaystyle{\sum_{k=1}^{50}}~k P[k]$

    you might also Wiki the binomial distribution for a simpler expression for the expected value, but the above is the definition of the expected value of this random variable.

    3) This is a more interesting problem. Here we have a binomial distribution with an unknown $n$

    We need to find $n$ such that $P[K\geq 10]=1-P[K \leq 9] = 0.95$

    You can either take a look at that Wiki page to see what the CDF is and try and solve it that way.

    Or you can set up some software to tabulate the probability as you vary $n$ and see how big you need to make $n$ to reach $p=0.95$
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