Question

according to the purpose of deciding if lots of computers that sends a factory are accepted, a procedure is done that consist on choosing 10 equips radomly from each lot and determinate the number of obsolete ones. Any lot is always refused if it's found 3 or more obsolete equips among 10 chosen lots. Suposse that number of items in every lot is big and each lot contains 5% obsolete equips. ¿what is the probability of accepting a lot? @agent0smith @mathstudent55 @ganeshie8

Please guys, help with this one (five points and your medal)

Answer 1

I don't know how to handle this one

Answer 2

No ideas in my mind now

Answer 3

@ganeshie8

Answer 4

the part that says 10 chosen lots is wrong... My bad, it should say 10 chosen

Answer 5

@ganeshie8 @agent0smith

Answer 6

Pretty confusingly written, but it sounds like a binomial problem.
http://fsweb.bainbridge.edu/dbyrd/statistics/binomial_formula.gif
N=10
p= 0.05 ("success" is getting an obsolete one)
q= 0.95

"lot is always refused if it's found 3 or more obsolete"
So it's accepted if there's less than or equal to 2 obsolete ones. You need:
\[P(X \le 2) = P(X=0) +P(X=1) +P(X=2) \]

Answer 7

mmmm

Answer 8

I should just replace the values given in link formula, shouldn't I?

Answer 9

I don't know how to apply that formula, sorry bro

Answer 10

Yes, i'll show you how to do P(X=0), you can do P(X=1) and P(X=2) the same way. Then just add them all together.

\[\large P(X=0) = \left(\begin{matrix}10 \\ 0\end{matrix}\right) 0.05^0 *0.95^{10}\]I assume you know how to do the combinations part; calculators have a button for it, too, or in the math menus.