An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn.Probability of selecting $3$ red and $2$ green from $50$ red and $50$ green ballsChoosing Balls: $3$ Green, $5$ Blue and $4$ red3 balls drawn from 1 urn - probability all same color (with/without replacement)an urn contains six ball of each of the three colors: red, blue, and green.An urn contains 5 green and 2 red balls.3 balls drawn from 1 urn - probability of getting exactly one colorA bag contains contains 20 blue marbles, 20 green marbles, and 20 red marblesDrawing 4 balls from an urn without replacement and a bonus ballAn urn containing $r$ red balls and $b$ blue balls.Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selectedAn urn contains 2 red balls and 3 blue balls.
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An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn.
Probability of selecting $3$ red and $2$ green from $50$ red and $50$ green ballsChoosing Balls: $3$ Green, $5$ Blue and $4$ red3 balls drawn from 1 urn - probability all same color (with/without replacement)an urn contains six ball of each of the three colors: red, blue, and green.An urn contains 5 green and 2 red balls.3 balls drawn from 1 urn - probability of getting exactly one colorA bag contains contains 20 blue marbles, 20 green marbles, and 20 red marblesDrawing 4 balls from an urn without replacement and a bonus ballAn urn containing $r$ red balls and $b$ blue balls.Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selectedAn urn contains 2 red balls and 3 blue balls.
$begingroup$
Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?
This is an exam question I got wrong. My answer was:
$frac5 choose 12 choose 29 choose 316 choose 6 $
I checked other questions, such as this one, and they approached it the same way. What am I missing?
probability combinatorics
asked Apr 30 at 17:41
juliodesajuliodesa
7517
$endgroup$
|
show 2 more comments
$begingroup$
Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?
This is an exam question I got wrong. My answer was:
$frac5 choose 12 choose 29 choose 316 choose 6 $
I checked other questions, such as this one, and they approached it the same way. What am I missing?
probability combinatorics
asked Apr 30 at 17:41
juliodesajuliodesa
7517
$endgroup$
$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
2
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50
|
show 2 more comments
$begingroup$
Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?
This is an exam question I got wrong. My answer was:
$frac5 choose 12 choose 29 choose 316 choose 6 $
I checked other questions, such as this one, and they approached it the same way. What am I missing?
probability combinatorics
asked Apr 30 at 17:41
juliodesajuliodesa
7517
$endgroup$
Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?
This is an exam question I got wrong. My answer was:
$frac5 choose 12 choose 29 choose 316 choose 6 $
I checked other questions, such as this one, and they approached it the same way. What am I missing?
probability combinatorics
probability combinatorics
asked Apr 30 at 17:41
juliodesajuliodesa
7517
asked Apr 30 at 17:41
juliodesajuliodesa
7517
asked Apr 30 at 17:41
juliodesajuliodesa
7517
asked Apr 30 at 17:41
juliodesajuliodesa
7517
asked Apr 30 at 17:41
juliodesajuliodesa
7517
7517
$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
2
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50
|
show 2 more comments
$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
2
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50
$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
2
2
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.
answered Apr 30 at 17:46
dxbdxb
3555
$endgroup$
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
add a comment |
$begingroup$
Since there is replacement there is no conditional probability, this means that the answer is simply:
$$frac6!3!2!1!p( colorredbullet cap colorbluebullet cap colorblue bullet capcolorgreenbullet capcolorgreenbulletcapcolorgreenbullet)=frac6!3!2!1!p(colorredbullet)p^2(colorbluebullet )p^3(colorgreenbullet colorgreen)=60left(frac516right)left(frac216right)^2 left(frac916right)^3=frac546751048576approx 5.21 % $$
Where the factor $frac6!3!2!1!$ indicates the number of permutations of the sequence of drawings.
:)
answered Apr 30 at 17:50
EurekaEureka
1,684117
$endgroup$
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.
answered Apr 30 at 17:46
dxbdxb
3555
$endgroup$
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
add a comment |
$begingroup$
Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.
answered Apr 30 at 17:46
dxbdxb
3555
$endgroup$
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
add a comment |
$begingroup$
Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.
answered Apr 30 at 17:46
dxbdxb
3555
$endgroup$
Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.
answered Apr 30 at 17:46
dxbdxb
3555
answered Apr 30 at 17:46
dxbdxb
3555
answered Apr 30 at 17:46
dxbdxb
3555
answered Apr 30 at 17:46
dxbdxb
3555
3555
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
add a comment |
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
$begingroup$
So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then?
$endgroup$
– juliodesa
Apr 30 at 17:47
2
2
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
$begingroup$
@juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution?
$endgroup$
– JMoravitz
Apr 30 at 17:48
add a comment |
$begingroup$
Since there is replacement there is no conditional probability, this means that the answer is simply:
$$frac6!3!2!1!p( colorredbullet cap colorbluebullet cap colorblue bullet capcolorgreenbullet capcolorgreenbulletcapcolorgreenbullet)=frac6!3!2!1!p(colorredbullet)p^2(colorbluebullet )p^3(colorgreenbullet colorgreen)=60left(frac516right)left(frac216right)^2 left(frac916right)^3=frac546751048576approx 5.21 % $$
Where the factor $frac6!3!2!1!$ indicates the number of permutations of the sequence of drawings.
:)
answered Apr 30 at 17:50
EurekaEureka
1,684117
$endgroup$
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
add a comment |
$begingroup$
Since there is replacement there is no conditional probability, this means that the answer is simply:
$$frac6!3!2!1!p( colorredbullet cap colorbluebullet cap colorblue bullet capcolorgreenbullet capcolorgreenbulletcapcolorgreenbullet)=frac6!3!2!1!p(colorredbullet)p^2(colorbluebullet )p^3(colorgreenbullet colorgreen)=60left(frac516right)left(frac216right)^2 left(frac916right)^3=frac546751048576approx 5.21 % $$
Where the factor $frac6!3!2!1!$ indicates the number of permutations of the sequence of drawings.
:)
answered Apr 30 at 17:50
EurekaEureka
1,684117
$endgroup$
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
add a comment |
$begingroup$
Since there is replacement there is no conditional probability, this means that the answer is simply:
$$frac6!3!2!1!p( colorredbullet cap colorbluebullet cap colorblue bullet capcolorgreenbullet capcolorgreenbulletcapcolorgreenbullet)=frac6!3!2!1!p(colorredbullet)p^2(colorbluebullet )p^3(colorgreenbullet colorgreen)=60left(frac516right)left(frac216right)^2 left(frac916right)^3=frac546751048576approx 5.21 % $$
Where the factor $frac6!3!2!1!$ indicates the number of permutations of the sequence of drawings.
:)
answered Apr 30 at 17:50
EurekaEureka
1,684117
$endgroup$
Since there is replacement there is no conditional probability, this means that the answer is simply:
$$frac6!3!2!1!p( colorredbullet cap colorbluebullet cap colorblue bullet capcolorgreenbullet capcolorgreenbulletcapcolorgreenbullet)=frac6!3!2!1!p(colorredbullet)p^2(colorbluebullet )p^3(colorgreenbullet colorgreen)=60left(frac516right)left(frac216right)^2 left(frac916right)^3=frac546751048576approx 5.21 % $$
Where the factor $frac6!3!2!1!$ indicates the number of permutations of the sequence of drawings.
:)
answered Apr 30 at 17:50
EurekaEureka
1,684117
edited Apr 30 at 17:55
answered Apr 30 at 17:50
EurekaEureka
1,684117
answered Apr 30 at 17:50
EurekaEureka
1,684117
answered Apr 30 at 17:50
EurekaEureka
1,684117
1,684117
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
add a comment |
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
1
1
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
My graphic is so lovely xd :)
$endgroup$
– Eureka
Apr 30 at 17:56
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
$begingroup$
@callculus I was too concentrated on the graphics lol
$endgroup$
– Eureka
Apr 30 at 17:58
add a comment |
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$begingroup$
You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible.
$endgroup$
– mihaild
Apr 30 at 17:44
$begingroup$
Why? I included "9 choose 3" in the fraction.
$endgroup$
– juliodesa
Apr 30 at 17:46
$begingroup$
Oops, a typo from my part: it should be $3$ blue balls.
$endgroup$
– mihaild
Apr 30 at 17:46
$begingroup$
But there are only 2 blue balls in the urn.
$endgroup$
– juliodesa
Apr 30 at 17:49
2
$begingroup$
And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after.
$endgroup$
– JMoravitz
Apr 30 at 17:50