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    This question is on Q4d on P17 of the Edexcel S2 Pearson Company textbook:

    A door to door canvasser tries to persuade people to have a certain type of double glazing installed. The probability that his canvassing at a house is successful is 0.05.

    Calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99.

    Help would be much appreciated. Thanks!
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    (Original post by Lady_Diamond)
    This question is on Q4d on P17 of the Edexcel S2 Pearson Company textbook:

    A door to door canvasser tries to persuade people to have a certain type of double glazing installed. The probability that his canvassing at a house is successful is 0.05.

    Calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99.

    Help would be much appreciated. Thanks!

    The way to go about this is to consider the complementary event.

    So instead of looking for the probablity of at least one success to exceed 0.99.

    Look for the probabilty of all houses canvassed failing to get double glazing installed to be less than 0.01.

    Can you take it from there?
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    Thanks. Ah I see. If possible, could you show the working as I am still struggling (but I understand your concept).
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    Do have a go yourself.

    I've put as far as creating the necessary equation to solve in the spoiler.

    Spoiler:
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    So you are looking to find the minimum n, such that

    P(no sales when canvassing n houses) < 0.01

    Assuming the canvassing of any house is independent of any other (which you must assume),

    P(no sales when canvassing n houses) = (P(no sales when canvassing one house))^n

    P(no sales when canvassing one house) = 1-0.05 = 0.95.

    So you just need to find the minimum n such that (0.95)^n < 0.01

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    Thank you.
 
 
 
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