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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 $100p =$ _____________.

@2019_Aspirant we can't do like this..

@Atul159, we cannot apply binomial distribution here because ;

There are three different criteria of binomial distributions described below which the binomial distributions need to fulfil.

• The number of the trial or the experiment must be fixed. As you can only figure out the probable chance of occurrence of success in a trail you should have a finite number of trials.
• Every trial is independent. None of your trials should affect the possibility of the next trial.
• The probability always stays the same and equal. The probability of success may be equal for more than one trial.

here though the no.of trials are fixed, but the trials are not independent of each other, because depending on our selection of one computer, if it is defective or not, we choose the next. Hence the problem follows Hypergeometric distribution.

Hope that clarifies your query!

Ref: NCERT

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 $100\times p =11.90$

Can this question be solved by binomial distribution. If yes then plss tell. As I am getting incorrect answer so want to know the approach.
reshown

Binomial distribution is applicable in case of independent events only.

In this problem,

1. You choose a system..either working or not working.( P[working]=4/10 and P[not working]=6/10 )

2. You choose next system..now the probability of choosing either working or not working computer depends on the choice that you made earlier in first step.

Thus one can say that events are not independent and thus binomial distribution can't be used here.

I hope it helps.

Correct me if am wrong.

Easier Method:-

We need to find probability of picking at least $three$ of the $four$ working computers. We have $10$ computers in total.

Sample space = No. of ways we can pick those 4 working computers among $10$ computers = $10C4 = 210$

Favourable cases = No. of ways of picking $>=3$ working computers

= No. of ways of picking $3$ working computers among those $4$ working computers and picking $1$ non-working computer among rest $6$ non-working computers + No. of ways of picking all $4$ working computers among all $4$ working computers

= $4C3*6C1 + 4C4 = 24+1=25$

Hence, $Probability (p)$ =  Favourable cases/Sample space= $25/210 = 0.1190476$

So, $100p = 0.1190476*100 = 11.904$ (Ans)

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

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

We can make use of Hypergeometric Distribution

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

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

$\therefore 100p=11.90$

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%