GATE CSE 2014 | Set 2 | Question: 1

Question

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 $100p =$ _____________.

Comments

 

Ref: NCERT

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Ans

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This can be solved using Hypergeometric Distribution method or Using simple probability counting method.

1) hypergeometric Distribution is solved by KUSHAGRA गुप्ता 

2) here W -> working and N--> not working ... we have combination WWWN + WWWW 

total number of unique aarangement of  WWWN = 4C3 and WWWW = 4C4 (order not matter here that's why we use combination) 

Final answer : 4C3*(4/10 * 3/9 *2/8 * 6/7) + 4C4 * (4/10 * 3/9 * 2/8 * 1/7)

Answer 1

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 $100\times p =11.90$

Comments

How can we say this??

 

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@Anant Singh_1 

it's told about the probability for working computers to be same, events are not same i.e the event of checking first computer at random and it's turn out to be working is different from the event where the working computer is the 4th computer.

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nice approach

Answer 2

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

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

Comments

op bhai

Answer 3

We can make use of Hypergeometric Distribution

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

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

$\therefore 100p=11.90$

Comments

Could you please tell me why i am wrong in this approach …

Number of favorable outcomes = 4C3 { chose 3 from 4 working computers } * 7C1{ chose 1 chair from the remaining  7 chairs } which consists of both working and not working computers

Number of outcomes = 10C4 .

so p = 7 * 8 / 210 which is not the answer..

Can you please explain what is wrong with my approach .

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Bro you sure you solving for this question ? 7C1{ chose 1 chair from the remaining  7 chairs } where is this coming from? 

Correct would be: 
Number of favorable outcomes = [4C3 { chose 3 from 4 working computers } * 6C1{ chose 1 computer from the remaining 6 non working computers} (here it is said that 4 are working properly and question is asked about at least 3 so we chose 3 from 4 and we choose remaining 1 from 6 non working)]  + [4C4 (because question asked at least 3 so we can choose all 4 working from 4 working computers)] = 24 + 1 = 25

now total sample space = 10C4 = 210 
p = 25/210 = 0.1190 
p*100 = 11.90

Answer 4

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 5

Here is the answer : 

Comments

If I use binomial distribution the answer was 17.92

Can anyone tell me why I can't use binomial distribution?

P = all are working + 3 are working out of 4

P = {4C4 * ( 4/10)^4 * (6/10)^0}  +   {4C3 * (4/10)^3 * (6/10)^1}

P = 0.0256 + 0.1536

P = 0.1792

 

Hence 100P = 17.92

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@Santi Singha If you select the first computer and it is working, there are now 3 working computers left out of 9. The probability of selecting another working computer has dropped from 0.4 to 3/9 = 0.333. In the binomial model, this probability would incorrectly remain 0.4 for all draws, leading to an overestimation. 

i.e. for example = choose 1 from working computer(from 4 working) now second computer we have to choose from 3 remaining computers. 1st choice 4/10, 2nd choice 3/9, and so on...  (But binomial won't change probabilities it will take 4/10 every time)

Each selection is independent, and the probability of success remains constant throughout the trials. However, in this case, the selection is without replacement, meaning the probabilities change after each draw. 
 

Answer 6

Ans