A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
A fastfood 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
anonymous
 one year ago
A fastfood 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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where are the answer so that mabye i can help u

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont have answer to try anything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry dude mabye i can find someone whose better hold on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A fastfood 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.)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1" 6363 percent pay nothing" Is a decimal point missing?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I am answering to this version of the question "A fastfood 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(13) = 1P(0) = 10.53^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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.)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Probability of getting nothing=0.53 Probability of getting fries = (10.47)/2=0.235 Probability of getting shake = (10.47)/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 giveaways, 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) = \(\left(\begin{matrix}3! \\ 1!1!1!\end{matrix}\right)(0.53^1)(0.235^1)(0.235^1)\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Given: (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)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.