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41 votes
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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 $100p =$ _____________.
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Best answer
56 votes
56 votes

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 $100\times p =11.90$

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

$\Rightarrow \left(\dfrac{1}{^{10}C_{4}}\right) + \dfrac{\left(^{4}C_{3} \times ^{6}C_{1}\right)}{^{10}C_{4}}$
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7 votes
7 votes

We can make use of Hypergeometric Distribution

This is particularly used in cases of sampling without replacement from a finite population.

$Required\ Probability\ (p)= \dfrac{^4C_{3}\times ^6C_{1}}{^{10}C_{4}}+\dfrac{^4C_{4}\times ^6C_{0}}{^{10}C_{4}}=\dfrac{24}{210}+\dfrac{1}{210}=\dfrac{25}{210}=0.1190$

$\therefore 100p=11.90$

5 votes
5 votes
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% 
Answer:

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