probability - a bag contains 10 blue marbles
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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
asked Dec 30 '18 at 15:03
swapnilswapnil
335
$endgroup$
add a comment |
$begingroup$
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
asked Dec 30 '18 at 15:03
swapnilswapnil
335
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 30 '18 at 15:03
swapnilswapnil
335
$endgroup$
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
asked Dec 30 '18 at 15:03
swapnilswapnil
335
asked Dec 30 '18 at 15:03
swapnilswapnil
335
asked Dec 30 '18 at 15:03
swapnilswapnil
335
asked Dec 30 '18 at 15:03
swapnilswapnil
335
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
$endgroup$
$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31
add a comment |
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$begingroup$
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
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
$endgroup$
$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31
add a comment |
$begingroup$
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
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
$endgroup$
$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31
add a comment |
$begingroup$
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
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
$endgroup$
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
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
answered Dec 30 '18 at 15:17
HenryHenry
100k481168
100k481168
$begingroup$
Thank you, I got it.
$endgroup$
– swapnil
Dec 30 '18 at 15:31
add a comment |
$begingroup$
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
add a comment |
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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