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Thread: Help on an extrmelye difficult probability question

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    Thumbs up Help on an extrmelye difficult probability question

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    If we have an empty room, and every week there is a 0.6 probability for a person to join, 0.2 probability for a person to leave, and 0.2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks?

    (A) 0.42
    (B) 0.55
    (C) 0.61
    (D) 0.67
    (E) 0.74

    Please explain how you do it, thanks!

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    Come on, nobody can help me?

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    Quote Originally Posted by cambridgedove View Post
    Come on, nobody can help me?
    How can there be a 20% chance of leaving the room if it is empty? Assuming the room can have negative occupancy, then by simulation the answer is 0.61.

    I don't know an easy way to calculate this. Note that the expected occupancy after 104 weeks is 104[(0.6)(1)+(0.2)(0)-(0.2)(1)] = 41.6

    Paul

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    1 out of 1 members found this post helpful. Good post? Yes | No
    If we have an empty room, and every week there is a 0.6 probability for a person to join, 0.2 probability for a person to leave, and 0.2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks?

    (A) 0.42
    (B) 0.55
    (C) 0.61
    (D) 0.67
    (E) 0.74

    Please explain how you do it, thanks!

    One way is to use the normal approximation. Remember if we have a large number of independent and identically distributed random variables and we sum them together we can approximate the distribution of the total sum with a normal distribution (central limit theorem). Here the number of people after 104 weeks is the sum of 104 independent and identically distributed random variables. Let Y be the random variable for the number of people added in a given week. Then E[Y] = {1*0.6+(-1)*0.2} =0.4. and Var[Y]=(1*0.6+1*0.2)-0.4^2=0.64. So the total sum after 104 weeks is approximated by a normal distribution N with E[N]=104*0.4=41.6 and Var[N]=104*0.64 = 66.56. So... now what's the probability we have more than 40 people? First we have to standardize the normal distribution by subtracting out the mean and divided by the standard deviation. (39-41.6)/sqrt(66.56) = -0.319. Find the probability that a standard normal random variable is greater than or equal to -0.319. For this... you need a cumulative normal distribution table. I'll look it up for you... and the result is that there is 0.38 probability that there are less than or equal to 39 people, so there is a 0.62 probability that there are 40 people or more. There could be an issue with the fact that you cannot leave the room if nobody is there. The problem as it is described is physically unrealistic as you could technically end up with a negative number of people in the room in the end.
    Last edited by eplownes; 03-01-2009 at 04:40 AM.

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