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Math Help - Converting from Individual probability to Household probability

  1. #1
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    Post Converting from Individual probability to Household probability

    I think this is basic, but I just can't figure it out. Any help would be appreciated...

    There is a 65% chance that each individual likes Product A.

    For a household of 2 individuals, what is the chance that someone in the household likes Product A?

    For a household of 3?

    I would appreciate the formula in addition to the answer if possible. Thank you very much for your help.
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  2. #2
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    Quote Originally Posted by RogBear62 View Post
    I think this is basic, but I just can't figure it out. Any help would be appreciated...

    There is a 65% chance that each individual likes Product A.

    For a household of 2 individuals, what is the chance that someone in the household likes Product A?

    For a household of 3?

    I would appreciate the formula in addition to the answer if possible. Thank you very much for your help.
    Let X be the random variable 'number of individuals in household who like product A'.

    X ~ Binomial(n = total number of individuals in household, p = 0.65).

    Calculate 1 - Pr(X = 0).
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  3. #3
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    Thank you for pointing me in the direction of binomial distributions!

    I think I figured out that in order for me to get the total probability for the household, I must use that formula for each member of the household, and add up the results. For instance, for a 3 person household my excel formula looks like:
    =BINOMDIST(1,3,.65,FALSE) +
    BINOMDIST(2,3,.65,FALSE) +
    BINOMDIST(3,3,.65,FALSE)

    (The format for the BINOMDIST formula in Excel is
    BINOMDIST(number_s, trials, probability_s, cumulative)
    Where number_s = the number of successes in trials,
    trials = the number of independent trials,
    Probability_s = probability of success,
    Cumulative - When FALSE returns probability that there are number_s successes.)

    This method returns these results for the given number of individuals in the household:
    1 = .65
    2 = .87
    3 = .95
    4 = .98
    5 = .99

    This "looks" right to me, but have I missed something? Thanks for your help.
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    Hi RogBear62,

    It is not necessary to add up the various binomial probabilities to get the answer. If you re-read Mr. F's solution, you will see he gave your a simpler method.

    Here is another route to the same answer, which maybe you will better understand. Let's say there are n people in the household, each of whom likes the product with probability 0.65. To find the probability that at least one person likes the product, consider the complementary problem: What is the probability that no one likes the product? This happens with probability 0.35^n. So the probability that at least one person likes the product is

    1 - 0.35^n.

    Note that this ASSUMES that each person's likes/dislikes are statistically independent. The same assumption is necessary for Mr. F's solution. In the Real World, this is unlikely.
    Last edited by awkward; May 29th 2010 at 07:34 AM. Reason: clarification
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