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Math Help - Please check my answers for probability hw (urgent)

  1. #1
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    Please check my answers for probability hw (urgent)

    Fistly, I'd like to say sorry about the text it is long. But I just want to check the answer. Thank you 4 ur time!

    (1). Printer cartridges are defective with probability p, independently of each other.

    (a) Cartridges are inspected one at a time. What is the probability that the fifth cartridge inspected is the second defective? What is the expected number of inspections required to find two defectives?

    (b) How many cartridges should you buy, to ensure a probability of at least 0.99 that at least one of them works?

    This is my answer:
    (a) Let X be the numbers of defective cartridges, we have X ~ NB(2,p)
    Then: P(X = 5) = 4.p^2(1-p)^3
    E(X) = 2/p

    (b) I let Y be the numbers of cartridges that work
    and let p' = 1-p be the probability of the cartridge that work
    n is the number of cartriges need to buy, then Y~Bin(n, p')
    We need P(Y>=1) >= 0.99
    after a long calculation, I got n>=(ln0.01/lnp)

    Is that right?. If I am wrong, please correct them

    2. There are 60 students in a class. 40 of them know that a random variable (r.v.) is a rule for allocating numbers to the outcomes of an experiment. These are ‘good’ students. A tutorial group contains 10 students, sampled independently from the class.
    a) What is the probability that the group contains exactly 8 good students?
    b) What is the probability that the group contains more good students than average?
    c) In fact, the group contains 5 good students. Their tutor asks to see all 10 students in turn. What is the probability that the sixth student seen is the third good one?


    This is my answer:
    Let X be the numbers of good student then X~H(10,40,60)
    a) P(X=8) = 0.1938

    b) E(X) = 20/3 = average ( say m)
    Then P(X> m) = P(X>=7) = 0.5594(4 d.p.)

    c) X~NB(3,0.5)
    P(X=6) = 0.156 (3 d.p.)

    Thanks again

    KN
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  2. #2
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    1) For the 5th cartridge to be the second defective one, two things must happen:
    A) out of the first four cartridges, 1 and only 1 must be defective.
    B) The 5th cartridge must be defective.
    We break down the problem as such.

    We have:

    p = probability of one cartridge being defective.

    For part A, we are concerned with the probability of getting exactly k defectivenesses (we can think of them as "successes" for pedagogy's sake) in n trials, where k=1 and n=4, obviously. This probability is given by the following Probability Mass Function (PMF):

     P_{n}(k) = \left( <br />
\begin{array}{c}<br />
n\\<br />
k\end{array}<br />
\right) <br />
p^k (1-p)^{n-k} =<br />
 \frac{n!}{k! (n-k)!} p^k (1-p)^{n-k}

    Now, plug in 4 for n, 1 for k, and p is p. What do we get?

     P_{4}(1) = \frac{4!}{1! (4-1)!} p^1 (1-p)^{4-1} <br />
= \frac{24}{6} p (1-p)^{3}<br />
= 4p(1-p)^{3}<br />
.

    Next, for part B, we know the probability of a cartridge being defective is p. So we multiply part A by part B, and we get the answer you got! Good job!
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