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Balls of 3 colours in a bag - probability of getting 3 same color balls in 3 out of 5 draws
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There are balls of 3 colours in a bag: 3 red, 4 green, 5 blue. Randomly perform 5 draws. In each draw, retrieve 3 balls from the bag at the same time . And place the balls back and perform next draw. What is the probability of getting 3 balls of same color in any 3 draws (means excluding 1,2,4,5 draws of the same color)?
probability combinatorics
asked Jul 19 at 22:03
techie11
104
 |Â
show 2 more comments
up vote
1
down vote
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There are balls of 3 colours in a bag: 3 red, 4 green, 5 blue. Randomly perform 5 draws. In each draw, retrieve 3 balls from the bag at the same time . And place the balls back and perform next draw. What is the probability of getting 3 balls of same color in any 3 draws (means excluding 1,2,4,5 draws of the same color)?
probability combinatorics
asked Jul 19 at 22:03
techie11
104
Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
1
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There are balls of 3 colours in a bag: 3 red, 4 green, 5 blue. Randomly perform 5 draws. In each draw, retrieve 3 balls from the bag at the same time . And place the balls back and perform next draw. What is the probability of getting 3 balls of same color in any 3 draws (means excluding 1,2,4,5 draws of the same color)?
probability combinatorics
asked Jul 19 at 22:03
techie11
104
There are balls of 3 colours in a bag: 3 red, 4 green, 5 blue. Randomly perform 5 draws. In each draw, retrieve 3 balls from the bag at the same time . And place the balls back and perform next draw. What is the probability of getting 3 balls of same color in any 3 draws (means excluding 1,2,4,5 draws of the same color)?
probability combinatorics
asked Jul 19 at 22:03
techie11
104
edited Jul 19 at 23:40
asked Jul 19 at 22:03
techie11
104
asked Jul 19 at 22:03
techie11
104
asked Jul 19 at 22:03
techie11
104
104
Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
1
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20
 |Â
show 2 more comments
Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
1
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20
Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
1
1
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).
First, find the probability that any given draw consists of just one color.
beginalign
P(textall one color)
& = P(textall red) + P(textall green) + P(textall blue) \
& = fracbinom33binom123
+ fracbinom43binom123
+ fracbinom53binom123
endalign
Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:
$$
P(textexactly $3$ of the $5$ three-ball draws are all one color)
= binom53 q^3 (1-q)^2
$$
answered Jul 19 at 22:24
Brian Tung
25.2k32453
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).
First, find the probability that any given draw consists of just one color.
beginalign
P(textall one color)
& = P(textall red) + P(textall green) + P(textall blue) \
& = fracbinom33binom123
+ fracbinom43binom123
+ fracbinom53binom123
endalign
Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:
$$
P(textexactly $3$ of the $5$ three-ball draws are all one color)
= binom53 q^3 (1-q)^2
$$
answered Jul 19 at 22:24
Brian Tung
25.2k32453
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
 |Â
show 1 more comment
up vote
2
down vote
accepted
Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).
First, find the probability that any given draw consists of just one color.
beginalign
P(textall one color)
& = P(textall red) + P(textall green) + P(textall blue) \
& = fracbinom33binom123
+ fracbinom43binom123
+ fracbinom53binom123
endalign
Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:
$$
P(textexactly $3$ of the $5$ three-ball draws are all one color)
= binom53 q^3 (1-q)^2
$$
answered Jul 19 at 22:24
Brian Tung
25.2k32453
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).
First, find the probability that any given draw consists of just one color.
beginalign
P(textall one color)
& = P(textall red) + P(textall green) + P(textall blue) \
& = fracbinom33binom123
+ fracbinom43binom123
+ fracbinom53binom123
endalign
Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:
$$
P(textexactly $3$ of the $5$ three-ball draws are all one color)
= binom53 q^3 (1-q)^2
$$
answered Jul 19 at 22:24
Brian Tung
25.2k32453
Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).
First, find the probability that any given draw consists of just one color.
beginalign
P(textall one color)
& = P(textall red) + P(textall green) + P(textall blue) \
& = fracbinom33binom123
+ fracbinom43binom123
+ fracbinom53binom123
endalign
Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:
$$
P(textexactly $3$ of the $5$ three-ball draws are all one color)
= binom53 q^3 (1-q)^2
$$
answered Jul 19 at 22:24
Brian Tung
25.2k32453
answered Jul 19 at 22:24
Brian Tung
25.2k32453
answered Jul 19 at 22:24
Brian Tung
25.2k32453
answered Jul 19 at 22:24
Brian Tung
25.2k32453
25.2k32453
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
 |Â
show 1 more comment
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains?
– techie11
Jul 19 at 22:39
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
@techie11: Are you asking that because of the Terence Tao question?
– Brian Tung
Jul 19 at 22:46
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
not actually. I just trying to understand the basic idea about counting :-)
– techie11
Jul 19 at 22:50
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@techie11 Clearly there are $binom 53, left(binom 33+binom 43+binom 53right)^3,left(binom123-(binom 33+binom 43+binom 53)right)^2$ favoured outcomes out of $binom123^5$ total.
– Graham Kemp
Jul 20 at 1:42
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
@Graham Kemp that is obvious. thanks.
– techie11
Jul 20 at 19:26
 |Â
show 1 more comment
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Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $binom 53$?
– lulu
Jul 19 at 22:08
1
Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation.
– lulu
Jul 19 at 22:11
I've edited the question and clarified. thanks.
– techie11
Jul 19 at 22:13
My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other).
– lulu
Jul 19 at 22:16
I did furthur editing to clarify.
– techie11
Jul 19 at 22:20