Mixing
We have $5$ red, $3$ blue and $2$ green balls. We take $3$ of them at random. What is the probability that...
$begingroup$
We have $5$ red, $3$ blue and $2$ green balls (balls of the same color are indistinguishable). We take $3$ of them at random. What is the probability that they are of different colors?
I have two ideas how to calculate and I think 2nd is right?
First: The number of all outcomes is $n = {10 choose 3} = 120$ and the number of good outcomes is $m = 5cdot 3cdot 2 $ so $$P= {mover n} = 0,25$$
Second: Probability of choosing $1$ red/blue/green ball is ${5over 10}$ /${3over 10}$/${2over 10}$. So choosing each color is ${3over 100}$, but we have $6$ different arrangements of good choices that give us $$P = 6cdot {3over 100} = 0,18$$
Now which one is correct? Or it is no one? If you offer new solution then please write down a sample space and probability function in your solution.
probability proof-verification alternative-proof
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
$endgroup$
add a comment |
$begingroup$
We have $5$ red, $3$ blue and $2$ green balls (balls of the same color are indistinguishable). We take $3$ of them at random. What is the probability that they are of different colors?
I have two ideas how to calculate and I think 2nd is right?
First: The number of all outcomes is $n = {10 choose 3} = 120$ and the number of good outcomes is $m = 5cdot 3cdot 2 $ so $$P= {mover n} = 0,25$$
Second: Probability of choosing $1$ red/blue/green ball is ${5over 10}$ /${3over 10}$/${2over 10}$. So choosing each color is ${3over 100}$, but we have $6$ different arrangements of good choices that give us $$P = 6cdot {3over 100} = 0,18$$
Now which one is correct? Or it is no one? If you offer new solution then please write down a sample space and probability function in your solution.
probability proof-verification alternative-proof
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
$endgroup$
$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39
add a comment |
$begingroup$
We have $5$ red, $3$ blue and $2$ green balls (balls of the same color are indistinguishable). We take $3$ of them at random. What is the probability that they are of different colors?
I have two ideas how to calculate and I think 2nd is right?
First: The number of all outcomes is $n = {10 choose 3} = 120$ and the number of good outcomes is $m = 5cdot 3cdot 2 $ so $$P= {mover n} = 0,25$$
Second: Probability of choosing $1$ red/blue/green ball is ${5over 10}$ /${3over 10}$/${2over 10}$. So choosing each color is ${3over 100}$, but we have $6$ different arrangements of good choices that give us $$P = 6cdot {3over 100} = 0,18$$
Now which one is correct? Or it is no one? If you offer new solution then please write down a sample space and probability function in your solution.
probability proof-verification alternative-proof
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
$endgroup$
We have $5$ red, $3$ blue and $2$ green balls (balls of the same color are indistinguishable). We take $3$ of them at random. What is the probability that they are of different colors?
I have two ideas how to calculate and I think 2nd is right?
First: The number of all outcomes is $n = {10 choose 3} = 120$ and the number of good outcomes is $m = 5cdot 3cdot 2 $ so $$P= {mover n} = 0,25$$
Second: Probability of choosing $1$ red/blue/green ball is ${5over 10}$ /${3over 10}$/${2over 10}$. So choosing each color is ${3over 100}$, but we have $6$ different arrangements of good choices that give us $$P = 6cdot {3over 100} = 0,18$$
Now which one is correct? Or it is no one? If you offer new solution then please write down a sample space and probability function in your solution.
probability proof-verification alternative-proof
probability proof-verification alternative-proof
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
edited Jan 19 at 14:32
greedoid
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
asked Oct 9 '18 at 22:28
greedoidgreedoid
45.1k1157112
45.1k1157112
$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39
add a comment |
$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39
$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39
add a comment |
2 Answers
2
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$begingroup$
Your first method is correct.
Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is
$$frac{5}{10} cdot frac{3}{9} cdot frac{2}{8} = frac{1}{24}$$
As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $frac{1}{4}$, which agrees with the answer obtained using your first method.
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
$begingroup$
I believe the first is correct and the second is flawed.
It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.
For instance the chance to pick red then blue then green is in fact $frac{5}{10}cdotfrac{3}{9}cdotfrac{2}{8}$
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Your first method is correct.
Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is
$$frac{5}{10} cdot frac{3}{9} cdot frac{2}{8} = frac{1}{24}$$
As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $frac{1}{4}$, which agrees with the answer obtained using your first method.
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
$begingroup$
Your first method is correct.
Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is
$$frac{5}{10} cdot frac{3}{9} cdot frac{2}{8} = frac{1}{24}$$
As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $frac{1}{4}$, which agrees with the answer obtained using your first method.
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
$begingroup$
Your first method is correct.
Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is
$$frac{5}{10} cdot frac{3}{9} cdot frac{2}{8} = frac{1}{24}$$
As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $frac{1}{4}$, which agrees with the answer obtained using your first method.
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
Your first method is correct.
Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is
$$frac{5}{10} cdot frac{3}{9} cdot frac{2}{8} = frac{1}{24}$$
As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $frac{1}{4}$, which agrees with the answer obtained using your first method.
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
edited Oct 9 '18 at 22:42
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
answered Oct 9 '18 at 22:39
N. F. TaussigN. F. Taussig
44.5k103357
44.5k103357
add a comment |
add a comment |
$begingroup$
I believe the first is correct and the second is flawed.
It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.
For instance the chance to pick red then blue then green is in fact $frac{5}{10}cdotfrac{3}{9}cdotfrac{2}{8}$
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
$endgroup$
add a comment |
$begingroup$
I believe the first is correct and the second is flawed.
It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.
For instance the chance to pick red then blue then green is in fact $frac{5}{10}cdotfrac{3}{9}cdotfrac{2}{8}$
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
$endgroup$
add a comment |
$begingroup$
I believe the first is correct and the second is flawed.
It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.
For instance the chance to pick red then blue then green is in fact $frac{5}{10}cdotfrac{3}{9}cdotfrac{2}{8}$
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
$endgroup$
I believe the first is correct and the second is flawed.
It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.
For instance the chance to pick red then blue then green is in fact $frac{5}{10}cdotfrac{3}{9}cdotfrac{2}{8}$
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
answered Oct 9 '18 at 22:37
YankoYanko
7,3301729
7,3301729
add a comment |
add a comment |
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$begingroup$
Before the problem can be solved, more information is needed: (1) are the balls selected with replacement or without replacement (i.e. after the first ball is selected, is it put back before the second ball is selected)? (2) what does "of a different color" mean, that the selected balls represent at least two different colors or that they represent all three colors?
$endgroup$
– user2661923
Oct 9 '18 at 22:38
$begingroup$
Without replacement and no two are of the same color
$endgroup$
– greedoid
Oct 9 '18 at 22:39