probability - a bag contains 10 blue marbles












0












$begingroup$


A bag contain 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. the probability that no two of the marbles drawn have the same color is____.



I considered different combinations of above scenario.



$
<R,B,G> <R,G,R>\
<R,G,B> <R,B,R>\
<B,R,G> <B,G,B>\
<B,G,R> <B,R,B>\
<G,R,B> <G,R,G>\
<G,B,R> <G,B,G>\
$



and my sample space will be = 3*3*3 -> each ball selection could be off 3 colors.
so probability becomes = $frac{12}{27}$ = $frac{4}{9}$
but it is not the correct answer.



I know I haven't counted no. of the given balls
So I though of this approach:
= $frac{12}{60_{C_3}}$



but no answer was still wrong. What should have been the correct way of solving it, and where I am making mistake?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    In e.g. scenario RGR two marbles of the same color are drawn.
    $endgroup$
    – drhab
    Dec 30 '18 at 15:17






  • 1




    $begingroup$
    math.stackexchange.com/questions/1848392/…
    $endgroup$
    – kludg
    Dec 30 '18 at 15:25






  • 1




    $begingroup$
    Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
    $endgroup$
    – David K
    Dec 30 '18 at 15:39
















0












$begingroup$


A bag contain 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. the probability that no two of the marbles drawn have the same color is____.



I considered different combinations of above scenario.



$
<R,B,G> <R,G,R>\
<R,G,B> <R,B,R>\
<B,R,G> <B,G,B>\
<B,G,R> <B,R,B>\
<G,R,B> <G,R,G>\
<G,B,R> <G,B,G>\
$



and my sample space will be = 3*3*3 -> each ball selection could be off 3 colors.
so probability becomes = $frac{12}{27}$ = $frac{4}{9}$
but it is not the correct answer.



I know I haven't counted no. of the given balls
So I though of this approach:
= $frac{12}{60_{C_3}}$



but no answer was still wrong. What should have been the correct way of solving it, and where I am making mistake?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    In e.g. scenario RGR two marbles of the same color are drawn.
    $endgroup$
    – drhab
    Dec 30 '18 at 15:17






  • 1




    $begingroup$
    math.stackexchange.com/questions/1848392/…
    $endgroup$
    – kludg
    Dec 30 '18 at 15:25






  • 1




    $begingroup$
    Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
    $endgroup$
    – David K
    Dec 30 '18 at 15:39














0












0








0





$begingroup$


A bag contain 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. the probability that no two of the marbles drawn have the same color is____.



I considered different combinations of above scenario.



$
<R,B,G> <R,G,R>\
<R,G,B> <R,B,R>\
<B,R,G> <B,G,B>\
<B,G,R> <B,R,B>\
<G,R,B> <G,R,G>\
<G,B,R> <G,B,G>\
$



and my sample space will be = 3*3*3 -> each ball selection could be off 3 colors.
so probability becomes = $frac{12}{27}$ = $frac{4}{9}$
but it is not the correct answer.



I know I haven't counted no. of the given balls
So I though of this approach:
= $frac{12}{60_{C_3}}$



but no answer was still wrong. What should have been the correct way of solving it, and where I am making mistake?






probability







share|cite|improve this question









$endgroup$




A bag contain 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. the probability that no two of the marbles drawn have the same color is____.



I considered different combinations of above scenario.



$
<R,B,G> <R,G,R>\
<R,G,B> <R,B,R>\
<B,R,G> <B,G,B>\
<B,G,R> <B,R,B>\
<G,R,B> <G,R,G>\
<G,B,R> <G,B,G>\
$



and my sample space will be = 3*3*3 -> each ball selection could be off 3 colors.
so probability becomes = $frac{12}{27}$ = $frac{4}{9}$
but it is not the correct answer.



I know I haven't counted no. of the given balls
So I though of this approach:
= $frac{12}{60_{C_3}}$



but no answer was still wrong. What should have been the correct way of solving it, and where I am making mistake?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 15:03









swapnilswapnil

335




335








  • 1




    $begingroup$
    In e.g. scenario RGR two marbles of the same color are drawn.
    $endgroup$
    – drhab
    Dec 30 '18 at 15:17






  • 1




    $begingroup$
    math.stackexchange.com/questions/1848392/…
    $endgroup$
    – kludg
    Dec 30 '18 at 15:25






  • 1




    $begingroup$
    Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
    $endgroup$
    – David K
    Dec 30 '18 at 15:39














  • 1




    $begingroup$
    In e.g. scenario RGR two marbles of the same color are drawn.
    $endgroup$
    – drhab
    Dec 30 '18 at 15:17






  • 1




    $begingroup$
    math.stackexchange.com/questions/1848392/…
    $endgroup$
    – kludg
    Dec 30 '18 at 15:25






  • 1




    $begingroup$
    Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
    $endgroup$
    – David K
    Dec 30 '18 at 15:39








1




1




$begingroup$
In e.g. scenario RGR two marbles of the same color are drawn.
$endgroup$
– drhab
Dec 30 '18 at 15:17




$begingroup$
In e.g. scenario RGR two marbles of the same color are drawn.
$endgroup$
– drhab
Dec 30 '18 at 15:17




1




1




$begingroup$
math.stackexchange.com/questions/1848392/…
$endgroup$
– kludg
Dec 30 '18 at 15:25




$begingroup$
math.stackexchange.com/questions/1848392/…
$endgroup$
– kludg
Dec 30 '18 at 15:25




1




1




$begingroup$
Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
$endgroup$
– David K
Dec 30 '18 at 15:39




$begingroup$
Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one.
$endgroup$
– David K
Dec 30 '18 at 15:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

Two issues:




  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s


  • The probability of drawing $R,B,G$ in that order is $frac{30}{60} times frac{10}{60} times frac{20}{60} = frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 times frac1{36}= frac16$, which I would expect to be the answer







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I got it.
    $endgroup$
    – swapnil
    Dec 30 '18 at 15:31












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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2












$begingroup$

Two issues:




  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s


  • The probability of drawing $R,B,G$ in that order is $frac{30}{60} times frac{10}{60} times frac{20}{60} = frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 times frac1{36}= frac16$, which I would expect to be the answer







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I got it.
    $endgroup$
    – swapnil
    Dec 30 '18 at 15:31
















2












$begingroup$

Two issues:




  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s


  • The probability of drawing $R,B,G$ in that order is $frac{30}{60} times frac{10}{60} times frac{20}{60} = frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 times frac1{36}= frac16$, which I would expect to be the answer







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, I got it.
    $endgroup$
    – swapnil
    Dec 30 '18 at 15:31














2












2








2





$begingroup$

Two issues:




  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s


  • The probability of drawing $R,B,G$ in that order is $frac{30}{60} times frac{10}{60} times frac{20}{60} = frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 times frac1{36}= frac16$, which I would expect to be the answer







share|cite|improve this answer









$endgroup$



Two issues:




  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s


  • The probability of drawing $R,B,G$ in that order is $frac{30}{60} times frac{10}{60} times frac{20}{60} = frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 times frac1{36}= frac16$, which I would expect to be the answer








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 15:17









HenryHenry

101k482170




101k482170












  • $begingroup$
    Thank you, I got it.
    $endgroup$
    – swapnil
    Dec 30 '18 at 15:31


















  • $begingroup$
    Thank you, I got it.
    $endgroup$
    – swapnil
    Dec 30 '18 at 15:31
















$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31




$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31


















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