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The security system at an IT office is composed of $10$ computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then $100$p = _____________.

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+11
Not working computers = 6
Working computers = 4

100*p(At least 3 are working) = 100*$(\frac{4C3*6C1}{10C4}$ + $\frac{4C4}{10C4})$ = 11.90
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thanx
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@Manu Thakur The story in the brackets can be more compressed.

$\frac{C(4,3)*C(7,1)}{C(10,4)}$

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Can we solve this problem using the binomial probability formula i.e ncx * p^x * q^n-x?

And if yes then how??

Initially $P$ (working computer) =$\dfrac{4}{10},$ $P$ (non-working computer) = $\dfrac{6}{10}.$

Case 1 : three computers are functional : There are $4$ sub-cases $WWWN, WWNW, WNWW, NWWW,$ where $W$ means working, $N$ means non-working, but $P(WWWN) = P(WWNW) = P(WNWW) = P(NWWW),$

because for example    $P(WWWN)=\dfrac{4}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{6}{7}=\dfrac{144}{5040}$

In all other $3$ sub-cases, we get same numerators and denominators (in different order), so total prob in this case is

$\dfrac{4\times 144}{5040} = \dfrac{576}{5040}$.

Case 2 : all $4$ are working

$P(WWWW)=\dfrac{4}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{1}{7}=\dfrac{24}{5040}$

$P$(atleast $3$ are working) =$\dfrac{600}{5040}$

So $100\times p =11.90$

edited
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once a working computer is picked, the probability of the next picked computer being working should go down rt?
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hmm...you are correct. I have edited the answer, please check if now it looks correct ?
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yes. Now its correct

+3
Hypergeometric distribution
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@Happy Mittal

Why did you multiply with 3 while choosing  3 functional computers?I mean why are we taking the subcases?

all are working $+ \;3$ Working and $1$ not working

$\Rightarrow \left(\dfrac{1}{^{10}C_{4}}\right) + \dfrac{\left(^{4}C_{3} \times ^{6}C_{1}\right)}{^{10}C_{4}}$
edited
Total ways to pick 4 computers = 10*9*8*7

Total ways that at least three computers are fine =
Total ways that all 4 are fine + Total ways any 3 are fine

Total ways that all 4 are fine = 4*3*2*1

Total ways three are fine = 1st is Not working and other 3 working +
2nd is Not working and other 3 working +
3rd is Not working and other 3 working +
4th is Not working and other 3 working +
= 6*4*3*2 + 4*6*3*2 + 4*3*6*2 + 4*3*2*6
= 6*4*3*2*4

The probability = Total ways that at least three computers are fine /
Total ways to pick 4 computers
=  (4*3*2*1 + 6*4*3*2*4) / (10*9*8*7)
= (4*3*2*25) / (10*9*8*7)
= 11.9% 

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