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?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
why isnt this question a binomial problem. oh because it is not with replacement
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
Not the answer you are looking for? Search for more explanations.
P ( X >= 2) = P (x=2) + P(x=3) + p(x=4)
P(x=2) = 9/12 *8/11*3/10*2/9 * 4!/(2!2!)
hero, do you agree?
Maybe this will help you:
easier to compute the probability that three are bad and subtract from one
probability that in the shipment of 4 you get the 3 bad ones is
That's what I got.
it says at least 2 , so
P( X > = 2 ) = P ( X = 2) + P ( X = 3) + P ( X = 4)
perl, does what I say make sense?
let me check
@satellite that is wrong
P ( x ≥ 2) = [ 9C2 * 3C2 + 9C3* 3C1+ 9C4 ] / 12C4 = 54/55
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
binomial is with replacment, this problem is without replacement
@inkyvoyd is probably wrong.
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
The Bernoulli trials are independent of one another.
The probability of success, p, remains the same from trial to trial.