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anonymous

  • one year ago

A​ fast-food chain randomly attaches coupons for prizes to the packages used to serve french fries. Most of the coupons say​ "Play again," but a few are winners. Of the​ coupons, 6363 percent pay​ nothing, with the rest evenly divided between​ "Win a free order of​ fries" and​ "Win a free​ sundae." Complete parts​ (a) through​ (c) below. ​(a) If each member of a family of three orders fries with her or his​ meal, what is the probability that someone in the family is a​ winner? The probability is

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  1. anonymous
    • one year ago
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    Where are the answer so that mabye i can help u

  2. anonymous
    • one year ago
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    try this

  3. anonymous
    • one year ago
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    What am i trying

  4. anonymous
    • one year ago
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    i dont have answer to try anything

  5. anonymous
    • one year ago
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    (a) If each member of a family of three orders fries with her or his​ meal, what is the probability that someone in the family is a​ winner? The probability is nothing. ​(Round to three decimal places as​ needed.)

  6. anonymous
    • one year ago
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    Sorry dude mabye i can find someone whose better hold on

  7. anonymous
    • one year ago
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    @purplemexican

  8. anonymous
    • one year ago
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    A​ fast-food chain randomly attaches coupons for prizes to the packages used to serve french fries. Most of the coupons say​ "Play again," but a few are winners. Of the​ coupons, 6363 percent pay​ nothing, with the rest evenly divided between​ "Win a free order of​ fries" and​ "Win a free​ sundae." Complete parts​ (a) through​ (c) below. ​(a) If each member of a family of three orders fries with her or his​ meal, what is the probability that someone in the family is a​ winner? The probability is nothing. ​(Round to three decimal places as​ needed.)

  9. mathmate
    • one year ago
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    " 6363 percent pay​ nothing" Is a decimal point missing?

  10. mathmate
    • one year ago
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    I am answering to this version of the question "A​ fast-food chain randomly attaches coupons for prizes to the packages used to serve french fries. Most of the coupons say​ "Play again," but a few are winners. Of the​ coupons, \(53\) percent pay​ nothing, with the rest evenly divided between​ "Win a free order of​ fries" and​ "Win a free​ sundae." Complete parts​ (a) through​ (c) below.​(a) If each member of a family of three orders fries with her or his​ meal, what is the probability that someone in the family is a​ winner?The probability is ​(Round to three decimal places as​ needed.)​(b) What is the probability that one member of the fam (a) Probability of NOT winning anything = 0.53 Probability of ALL three members of the family win nothing P(0)= (0.53)^3 by the multiplication rule of a three step experiment. so probability of a member winning something is P(1-3) = 1-P(0) = 1-0.53^3

  11. anonymous
    • one year ago
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    ​(b) What is the probability that one member of the family gets a free order of fries and another gets the​ sundae? The third wins nothing. The probability is ​(Round to four decimal places as​ needed.)

  12. mathmate
    • one year ago
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    Probability of getting nothing=0.53 Probability of getting fries = (1-0.47)/2=0.235 Probability of getting shake = (1-0.47)/2=0.235 Again, use the multiplication rule, Probability A gets nothing, B gets fries, C gets shake=0.53*0.235*0.235 But there are 3! ways to permutate the give-aways, so P(N,F,S) in any order = 3!(0.53)(0.235)(0.235) Alternatively use the multinomial theorem to get the same results: P(N,F,S) = \(\left(\begin{matrix}3! \\ 1!1!1!\end{matrix}\right)(0.53^1)(0.235^1)(0.235^1)\)

  13. mathmate
    • one year ago
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    Given: (c) The fries normally cost​ $1 and the sundae​ $2. What are the chances of the family winning​ $5 or more in​ prizes? The probability is ​(Round to five decimal places as​ needed.) pl help me out There are two cases: Two sundaes and 1 friesc = 2+2+1=$5 or three sundaes, value = 3*$2 = $6. Calculate the probability of each of the two cases and add them (since they are mutually exclusive, so the sum of probabilities equals the probability of one case or the other)

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