perl
  • perl
A shipment of 12 microwave ovens contains 3 defective units. A vending company has ordered 4 of these units, and because all are packaged identically, the selection will be at random. What is the probability that at least 2 units are good?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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perl
  • perl
why isnt this question a binomial problem. oh because it is not with replacement
Callisto
  • Callisto
I'm not sure for this .... probability that at least 2 units are good = 1 - P(3 are bad) = 1- 3/12 x 2/12 x 1/12 x 9/12 = 1 - 1/384 I'm really not good at Probability
perl
  • perl
ok i give answer

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More answers

perl
  • perl
P ( X >= 2) = P (x=2) + P(x=3) + p(x=4)
perl
  • perl
P(x=2) = 9/12 *8/11*3/10*2/9 * 4!/(2!2!)
perl
  • perl
hero, do you agree?
anonymous
  • anonymous
Maybe this will help you: http://answers.yahoo.com/question/index?qid=20110224214712AAmLuiv
anonymous
  • anonymous
easier to compute the probability that three are bad and subtract from one
anonymous
  • anonymous
probability that in the shipment of 4 you get the 3 bad ones is \[\frac{\dbinom{9}{4}}{\dbinom{12}{4}}\]
inkyvoyd
  • inkyvoyd
1-(3/12)*(2/11)*(1/10)*(9/9)*4.
inkyvoyd
  • inkyvoyd
That's what I got.
perl
  • perl
it says at least 2 , so P( X > = 2 ) = P ( X = 2) + P ( X = 3) + P ( X = 4)
inkyvoyd
  • inkyvoyd
perl, does what I say make sense?
perl
  • perl
let me check
perl
  • perl
@satellite that is wrong
anonymous
  • anonymous
P ( x ≥ 2) = [ 9C2 * 3C2 + 9C3* 3C1+ 9C4 ] / 12C4 = 54/55
Directrix
  • Directrix
Probability of a defective oven = 3/12 = 1/4 = .25 Probability of a non-defective oven = .75 P( 2 Defects) = C(4,2) * (.25)^2 * (.75)^2 = .2109 P( 3 Defects) = C(4,3) * (.25)^3 * (.75)^1 = .0469 P(4 Defects) = C(4,4) * (.25)^4 * (.75)^0 = .0039 Probability of at least 2 defective ovens = .2109 + .0469 + .0039 = .2617 http://openstudy.com/study#/updates/4f7935efe4b0ddcbb89ea1e9
perl
  • perl
binomial is with replacment, this problem is without replacement
inkyvoyd
  • inkyvoyd
@inkyvoyd is probably wrong.
Directrix
  • Directrix
A binomial model is characterized by trials which either end in success (heads) or failure (tails). These are sometimes called Bernoulli trials . Suppose we have n Bernoulli trials and p is the probability of success on a trial. Then this is a binomial model if 1. The Bernoulli trials are independent of one another. 2. The probability of success, p, remains the same from trial to trial. http://www.stat.wmich.edu/s160/book/node33.html
perl
  • perl
this is no binomial problem

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