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S2 probability question (how do i do it?) watch

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    4 (a) State clearly the conditions under which it is appropriate to assume that a random variable has a binomial distribution
    (1) fixed number of trials, (2) each trial is either a success or failure, (3) the probability of a success/failure is constant, (4) the events are independent

    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

    (b) Find the probability that he wll have at least 2 successes out of the first 10 houses he canvasses

    P(X>=2) = 1 - P(X=<2), then work out from cumulative binomial table (is this right?)

    (c) Find the number of houses he should canvass per day in order to average 3 successes per day.

    how do i do this part? :O any help would be much appreciated, thanks
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    For (b) that's not right. Since you're using discrete data, you can use either \mathbb{P}(X \ge 2) = 1 - \mathbb{P}(X &lt; 2) or \mathbb{P}(X \ge 2) = 1 - \mathbb{P}(X \le 1).

    For (c), you need to think about the expectation. That is, you need to find the number of houses for which \mathbb{E}(X) = 3.
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    (Original post by nuodai)
    For (b) that's not right. Since you're using discrete data, you can use either \mathbb{P}(X \ge 2) = 1 - \mathbb{P}(X &lt; 2) or \mathbb{P}(X \ge 2) = 1 - \mathbb{P}(X \le 1).

    For (c), you need to think about the expectation. That is, you need to find the number of houses for which \mathbb{E}(X) = 3.
    ah yeah i see what you mean about (b)

    and for (c), expectation = np right? so 3 = 0.05n, n = 3/0.05. is this the correct answer then?

    p.s. part (d) says: Calculte the least number of houses that he must canvass in order that the probability of him getting at least one success exceeds 0.99

    i don't understand this at all, care to push me in the right direction? would be much appreciated
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    (Original post by doin' your mum)
    ah yeah i see what you mean about (b)

    and for (c), expectation = np right? so 3 = 0.05n, n = 3/0.05. is this the correct answer then?
    Yup.

    (Original post by doin' your mum)
    p.s. part (d) says: Calculte the least number of houses that he must canvass in order that the probability of him getting at least one success exceeds 0.99

    i don't understand this at all, care to push me in the right direction? would be much appreciated
    This is finding the range of values of n such that P(X \ge 1) &gt; 0.99, and then the answer is the smallest value of n in this range. Using the right formula in the inequality P(X \ge 1) &gt; 0.99 and rearranging will give you an answer in the form n &gt; \text{something} -- bear in mind that if you have say n &gt; 32.61 (picked at random, not actually the answer) then the answer would be 33.

    The hardest part in most S2 problems is converting the word problem into a mathematical formula for you to solve... once you've got that down you'll be able to do these without any problems. Can you see where I got P(X \ge 1) &gt; 0.99 from?
 
 
 
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