probability that the system is deemed functional












2












$begingroup$


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 $100p=$




My Approach



I Know that this questio can be solved by -:



$P text{(Atleast 3 working=All four working +3 working and 1 not working}$



$$frac{1}{binom{10}{4}}+frac{binom{4}{3} times binom{6}{1}}{binom{10}{4}}$$



But i have doubt in another approach ,using Binomial distribution



$P(text{working})=frac{4}{10}$



$P text{(Atleast 3 working)=All four working +3 working and 1 not working}$



$$=binom{10}{3} times (frac{4}{10})^{3} times (frac{6}{10})^{7}+binom{10}{4} times (frac{4}{10})^{4} times (frac{6}{10})^{6}$$



In both case i am getting different answer .Why ?
Please help.I am totally stucked






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    please help me out
    $endgroup$
    – laura
    Jun 14 '18 at 11:23
















2












$begingroup$


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 $100p=$




My Approach



I Know that this questio can be solved by -:



$P text{(Atleast 3 working=All four working +3 working and 1 not working}$



$$frac{1}{binom{10}{4}}+frac{binom{4}{3} times binom{6}{1}}{binom{10}{4}}$$



But i have doubt in another approach ,using Binomial distribution



$P(text{working})=frac{4}{10}$



$P text{(Atleast 3 working)=All four working +3 working and 1 not working}$



$$=binom{10}{3} times (frac{4}{10})^{3} times (frac{6}{10})^{7}+binom{10}{4} times (frac{4}{10})^{4} times (frac{6}{10})^{6}$$



In both case i am getting different answer .Why ?
Please help.I am totally stucked






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    please help me out
    $endgroup$
    – laura
    Jun 14 '18 at 11:23














2












2








2





$begingroup$


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 $100p=$




My Approach



I Know that this questio can be solved by -:



$P text{(Atleast 3 working=All four working +3 working and 1 not working}$



$$frac{1}{binom{10}{4}}+frac{binom{4}{3} times binom{6}{1}}{binom{10}{4}}$$



But i have doubt in another approach ,using Binomial distribution



$P(text{working})=frac{4}{10}$



$P text{(Atleast 3 working)=All four working +3 working and 1 not working}$



$$=binom{10}{3} times (frac{4}{10})^{3} times (frac{6}{10})^{7}+binom{10}{4} times (frac{4}{10})^{4} times (frac{6}{10})^{6}$$



In both case i am getting different answer .Why ?
Please help.I am totally stucked






probability







share|cite|improve this question











$endgroup$




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 $100p=$




My Approach



I Know that this questio can be solved by -:



$P text{(Atleast 3 working=All four working +3 working and 1 not working}$



$$frac{1}{binom{10}{4}}+frac{binom{4}{3} times binom{6}{1}}{binom{10}{4}}$$



But i have doubt in another approach ,using Binomial distribution



$P(text{working})=frac{4}{10}$



$P text{(Atleast 3 working)=All four working +3 working and 1 not working}$



$$=binom{10}{3} times (frac{4}{10})^{3} times (frac{6}{10})^{7}+binom{10}{4} times (frac{4}{10})^{4} times (frac{6}{10})^{6}$$



In both case i am getting different answer .Why ?
Please help.I am totally stucked






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 14 '18 at 6:48







laura

















asked Jun 13 '18 at 8:00









lauralaura

1,2691824




1,2691824












  • $begingroup$
    please help me out
    $endgroup$
    – laura
    Jun 14 '18 at 11:23


















  • $begingroup$
    please help me out
    $endgroup$
    – laura
    Jun 14 '18 at 11:23
















$begingroup$
please help me out
$endgroup$
– laura
Jun 14 '18 at 11:23




$begingroup$
please help me out
$endgroup$
– laura
Jun 14 '18 at 11:23










2 Answers
2






active

oldest

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0












$begingroup$

The probability of having a working computer is $p=0.4$.



The experiment of the problem indeed follows the Binomial Distribution, as we want to express the number of successes ($3$ working computers) in $4$ trials.



The p.d.f. of the Binomial Distribution for our specific experiment, is :



$$mathbb{P}(p|n,k)=binom{n}{k}(1-0.4)^{n-k}cdot 0.4^k = binom{n}{k}0.6^{n-k}cdot 0.4^k$$



Now as mentioned, we want $3$ successes in $4$ trials, thus $n=4$ and $k=3$ and :



$$mathbb{P}(p=0.4|n=4,k=3)=binom{4}{3}0.6^{4-3}cdot0.4^3$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
    $endgroup$
    – laura
    Jun 13 '18 at 9:32



















0












$begingroup$

The binomial distribution could be used if computers where selected with replacement. Without replacement you have to use the hypergeometric distribution, and you obtain the same result you mention.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    The probability of having a working computer is $p=0.4$.



    The experiment of the problem indeed follows the Binomial Distribution, as we want to express the number of successes ($3$ working computers) in $4$ trials.



    The p.d.f. of the Binomial Distribution for our specific experiment, is :



    $$mathbb{P}(p|n,k)=binom{n}{k}(1-0.4)^{n-k}cdot 0.4^k = binom{n}{k}0.6^{n-k}cdot 0.4^k$$



    Now as mentioned, we want $3$ successes in $4$ trials, thus $n=4$ and $k=3$ and :



    $$mathbb{P}(p=0.4|n=4,k=3)=binom{4}{3}0.6^{4-3}cdot0.4^3$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
      $endgroup$
      – laura
      Jun 13 '18 at 9:32
















    0












    $begingroup$

    The probability of having a working computer is $p=0.4$.



    The experiment of the problem indeed follows the Binomial Distribution, as we want to express the number of successes ($3$ working computers) in $4$ trials.



    The p.d.f. of the Binomial Distribution for our specific experiment, is :



    $$mathbb{P}(p|n,k)=binom{n}{k}(1-0.4)^{n-k}cdot 0.4^k = binom{n}{k}0.6^{n-k}cdot 0.4^k$$



    Now as mentioned, we want $3$ successes in $4$ trials, thus $n=4$ and $k=3$ and :



    $$mathbb{P}(p=0.4|n=4,k=3)=binom{4}{3}0.6^{4-3}cdot0.4^3$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
      $endgroup$
      – laura
      Jun 13 '18 at 9:32














    0












    0








    0





    $begingroup$

    The probability of having a working computer is $p=0.4$.



    The experiment of the problem indeed follows the Binomial Distribution, as we want to express the number of successes ($3$ working computers) in $4$ trials.



    The p.d.f. of the Binomial Distribution for our specific experiment, is :



    $$mathbb{P}(p|n,k)=binom{n}{k}(1-0.4)^{n-k}cdot 0.4^k = binom{n}{k}0.6^{n-k}cdot 0.4^k$$



    Now as mentioned, we want $3$ successes in $4$ trials, thus $n=4$ and $k=3$ and :



    $$mathbb{P}(p=0.4|n=4,k=3)=binom{4}{3}0.6^{4-3}cdot0.4^3$$






    share|cite|improve this answer









    $endgroup$



    The probability of having a working computer is $p=0.4$.



    The experiment of the problem indeed follows the Binomial Distribution, as we want to express the number of successes ($3$ working computers) in $4$ trials.



    The p.d.f. of the Binomial Distribution for our specific experiment, is :



    $$mathbb{P}(p|n,k)=binom{n}{k}(1-0.4)^{n-k}cdot 0.4^k = binom{n}{k}0.6^{n-k}cdot 0.4^k$$



    Now as mentioned, we want $3$ successes in $4$ trials, thus $n=4$ and $k=3$ and :



    $$mathbb{P}(p=0.4|n=4,k=3)=binom{4}{3}0.6^{4-3}cdot0.4^3$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 13 '18 at 8:30









    RebellosRebellos

    14.6k31247




    14.6k31247












    • $begingroup$
      So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
      $endgroup$
      – laura
      Jun 13 '18 at 9:32


















    • $begingroup$
      So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
      $endgroup$
      – laura
      Jun 13 '18 at 9:32
















    $begingroup$
    So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
    $endgroup$
    – laura
    Jun 13 '18 at 9:32




    $begingroup$
    So the answer should be $=binom{4}{3} times (.4)^3 times (.6)^{1}+binom{4}{4} times (.4)^4 times (.6)^{0}=0.1536+0.0256=0.1792$ But the answer is $0.119$
    $endgroup$
    – laura
    Jun 13 '18 at 9:32











    0












    $begingroup$

    The binomial distribution could be used if computers where selected with replacement. Without replacement you have to use the hypergeometric distribution, and you obtain the same result you mention.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The binomial distribution could be used if computers where selected with replacement. Without replacement you have to use the hypergeometric distribution, and you obtain the same result you mention.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The binomial distribution could be used if computers where selected with replacement. Without replacement you have to use the hypergeometric distribution, and you obtain the same result you mention.






        share|cite|improve this answer









        $endgroup$



        The binomial distribution could be used if computers where selected with replacement. Without replacement you have to use the hypergeometric distribution, and you obtain the same result you mention.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 11:10









        JordiJordi

        31118




        31118






























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