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Yesterday, 19:00

You collect baseball cards and buy sealed packs from the grocery store and clearly have no idea which cards are inside (nor if they will be valuable). Suppose 1 in every 250 cards is valuable and a package contains 10 cards which are selected independently and randomly at the factory.

(a) If you purchase a package, what is the probability that there are no valuable cards? t you pucha a packae, wst s the probbiiythat thee is a east I valuable curd

(c) What is the expected number of valuable cards in the package?

(d) Suppose you buy 5 packages, what is the probability that none of these packages have valuable cards in them?

(e) Suppose you keep buying packages until you get 1 valuable card. What is the probability that you buy 2 packages?

(f) Suppose you desire to have 3 valuable cards in your collection. What is the probability that it takes 20 packages to accomplish this? On average, how many packages would it take to accomplish this?

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  1. Yesterday, 19:48
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    (a) 0.9607

    (c) 0.04

    (d) 0.8184

    (e) 0.074

    (f) 0.0853/e^0.8;

    750 cards

    Step-by-step explanation:

    Let the probability of success of a valuable card be p

    p = 1/250 = 0.004,

    q = 1-p = (249/250)

    (a) There are 10 cards in one pack

    The probability that there are no valuable card is given by:

    Pr (X = 0) = 10C0 * (1/250) ^0 * (249/250) ^10

    10C0 = 10 combination 0 = 10!/10!0!

    =1

    Pr (X = 0) = (249/250) ^10

    Pr (X=0) = 0.9607

    (c) Expected number of valuable card is given by:

    E (X) = np, n = 10, p = 1/250

    E (X) = 10 * 1/250 = 1/25 = 0.04

    (d) 5 packages means 5 * 10 = 50 cards

    Pr (X=0) = 50C0 * (1/250) ^0 * (249/250) ^50

    50C0 = 50!/50! * 0!=1

    Pr (X=0) = (249/250) ^50

    Pr (X = 0) = (0.996) ^50

    Pr (X=0) = 0.8184

    (e) For this case we use the Binomial Distribution

    2 packages 2 * 10 = 20 cards

    n = 20, x = 1

    We use:

    P (X=1) = 20C1 * (1/250) ^ 1 * (249/250) ^19

    Pr (X=1) = 20C1 * (1/250) ^1 * (249/250) ^19

    Pr (X=1) = 20 * (1/250) ¹ * (249/250) ^19

    Pr (X = 1) = 0.074

    (f) 20 packages = 200 cards

    For this we apply Poisson distribution

    P (X=3) = (e ^-h * h ^x) / x!

    Where h = np = 200/250 = 0.8

    P (X=3) = e^-0.8 * (0.8) ^3 / 3!

    P (X=3) = 0.991 * 0.512/6

    P (X=3) = 0.0846

    3 = n p

    n = 3/p = 3 / 0.04

    n = 750 cards
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