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A company recive 35 packages with light bulbs. In each pack there is 10 light bulbs, 2 of the light bulbs in each package has an error (witch cant bee seen). From each pack there is 3 diffrent bulbs picked out
What are the estimated chance that we will pick more than 0.35 light bulbs with the error?
1. 0,9471
2. 0,0078
3. 0,6110
4. 0,9922

witch cant bee seen

what ever dude i need the anwer

None. I use LED.

good 4 you my man

Man, i wont do ur homework 😎

i have to pass this test to take finals and i cant do this excercise cuz i play to much cs go ... and didnt go to class for this lecture...

And we should reward such behaviour by giving you the answer?

I guess thats ur problem right

you can look up how to solve this in 5 minutes on wikipedia

I can help with calculus. not this tho

binomials bruh

You'll fail the test, hf

Since all packages are the same, the number of packages doesn't matter anyway
And none of these 4 answers is the right one

Correct if my iq is low, but picking more than 0.35 💡 means at least one right ? Cause if so, it’s the opposite of not picking any which is ez mens))😎

maybe he means the chance that 35% of picked light bulbs has the error? idk. seems like standard statistics question. should be easily solved with the book.

Yes maybe. But still ez mens 😎

it isnt a standard problem.

its a variation on a standard problem, ive done like a dozen versions of it in probability & statistics class
in the standard version the total error rate is given instead of the condition 2/10 broken bulbs per box but you still pick them random so its the same as 20% error rate
also usually the question is what is the chance there will be 1 or more broken bulbs in a box

we're not helping with ur maths homework

i need a calculator

"we will pick more than 0.35 light bulbs" It must be 35%
You see the question is wrongly stated.

Google binomial chance

0.99477019784810140000
its a hypergeometrical distribution