Question

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?

Answer 1

why isnt this question a binomial problem. oh because it is not with replacement

Answer 2

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

Answer 3

ok i give answer

Answer 4

P ( X >= 2) = P (x=2) + P(x=3) + p(x=4)

Answer 5

P(x=2) = 9/12 *8/11*3/10*2/9 * 4!/(2!2!)

Answer 6

hero, do you agree?

Answer 7

Maybe this will help you:
http://answers.yahoo.com/question/index?qid=20110224214712AAmLuiv

Answer 8

easier to compute the probability that three are bad and subtract from one

Answer 9

probability that in the shipment of 4 you get the 3 bad ones is
\[\frac{\dbinom{9}{4}}{\dbinom{12}{4}}\]

Answer 10

1-(3/12)*(2/11)*(1/10)*(9/9)*4.

Answer 11

That's what I got.

Answer 12

it says at least 2 , so
P( X > = 2 ) = P ( X = 2) + P ( X = 3) + P ( X = 4)

Answer 13

perl, does what I say make sense?

Answer 14

let me check

Answer 15

@satellite that is wrong

Answer 16

P ( x ≥ 2) = [ 9C2 * 3C2 + 9C3* 3C1+ 9C4 ] / 12C4 = 54/55

Answer 17

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

Answer 18

binomial is with replacment, this problem is without replacement

Answer 19

@inkyvoyd is probably wrong.

Answer 20

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

Answer 21

this is no binomial problem