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Maximazing multinomial distribution
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I have an exercise in probability theory, which I can't solve:
There are 3 factories A B C, which produce 3 types of light bulbs. Factory A / B / C makes 40 / 20 / 40 percent of whole light bulbs. The probability of manufacturing first type of light bulb in factory A / B / C is 0.6 / 0.3 / 0.5. The probability of manufacturing second type of light bulb in factory A/ B / C is 0.3/0.4/0.2. The probability of manufacturing third type of light bulb in factory A / B / C is 0.1/0.3/0.3.
We buy 6 light bulbs. What is the most probable set of light bulbs (number of each type)? What is its probability?
So I figured out it's multinomial distribution with parameteres 6 and probabilities $p_1=0.5$, $p_2=0,22$, $p_3=0.28$, where $p_i$ is probability of buying a light bulb of $i-th$ type.
Let $X_i$ denote a number of light bulbs of $i-th$ type in the set of 6 light bulbs we bought. To solve first problem we want to maximaze $P(X_1=k_1, X_2=k_2, X_3=k_3) = frac6!k_1!cdot k_2! cdot k_3!0.5^k_1 cdot 0.22^k_2 cdot 0.28^k_3$
So we can write a function $f(x,y,z)=frac6!x!cdot y! cdot z! 0.5^xcdot 0.22^ycdot 0.28^z$. Since sum of all variables must be 6, we can reduce one variable. Then we want to maximaze function $g(x,y)=frac6!x!cdot y!cdot (6-x-y)!0.5^xcdot 0.22^ycdot 0.28^6-x-y$.
I don't how to do it because of the factorials. And also $k_i$ are natural numbers so i doubt that this aproach is good. I think you can examine this formula with sequences, but there is a lot of work to do then. I think there should be a theorem, which I don't know, that would help, because it was for an exam on some not mathematical (economy sth) studies and they have a small amount of math classes there.
probability
asked 2 days ago
Derm
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I have an exercise in probability theory, which I can't solve:
There are 3 factories A B C, which produce 3 types of light bulbs. Factory A / B / C makes 40 / 20 / 40 percent of whole light bulbs. The probability of manufacturing first type of light bulb in factory A / B / C is 0.6 / 0.3 / 0.5. The probability of manufacturing second type of light bulb in factory A/ B / C is 0.3/0.4/0.2. The probability of manufacturing third type of light bulb in factory A / B / C is 0.1/0.3/0.3.
We buy 6 light bulbs. What is the most probable set of light bulbs (number of each type)? What is its probability?
So I figured out it's multinomial distribution with parameteres 6 and probabilities $p_1=0.5$, $p_2=0,22$, $p_3=0.28$, where $p_i$ is probability of buying a light bulb of $i-th$ type.
Let $X_i$ denote a number of light bulbs of $i-th$ type in the set of 6 light bulbs we bought. To solve first problem we want to maximaze $P(X_1=k_1, X_2=k_2, X_3=k_3) = frac6!k_1!cdot k_2! cdot k_3!0.5^k_1 cdot 0.22^k_2 cdot 0.28^k_3$
So we can write a function $f(x,y,z)=frac6!x!cdot y! cdot z! 0.5^xcdot 0.22^ycdot 0.28^z$. Since sum of all variables must be 6, we can reduce one variable. Then we want to maximaze function $g(x,y)=frac6!x!cdot y!cdot (6-x-y)!0.5^xcdot 0.22^ycdot 0.28^6-x-y$.
I don't how to do it because of the factorials. And also $k_i$ are natural numbers so i doubt that this aproach is good. I think you can examine this formula with sequences, but there is a lot of work to do then. I think there should be a theorem, which I don't know, that would help, because it was for an exam on some not mathematical (economy sth) studies and they have a small amount of math classes there.
probability
asked 2 days ago
Derm
11
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago
add a comment |Â
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up vote
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I have an exercise in probability theory, which I can't solve:
There are 3 factories A B C, which produce 3 types of light bulbs. Factory A / B / C makes 40 / 20 / 40 percent of whole light bulbs. The probability of manufacturing first type of light bulb in factory A / B / C is 0.6 / 0.3 / 0.5. The probability of manufacturing second type of light bulb in factory A/ B / C is 0.3/0.4/0.2. The probability of manufacturing third type of light bulb in factory A / B / C is 0.1/0.3/0.3.
We buy 6 light bulbs. What is the most probable set of light bulbs (number of each type)? What is its probability?
So I figured out it's multinomial distribution with parameteres 6 and probabilities $p_1=0.5$, $p_2=0,22$, $p_3=0.28$, where $p_i$ is probability of buying a light bulb of $i-th$ type.
Let $X_i$ denote a number of light bulbs of $i-th$ type in the set of 6 light bulbs we bought. To solve first problem we want to maximaze $P(X_1=k_1, X_2=k_2, X_3=k_3) = frac6!k_1!cdot k_2! cdot k_3!0.5^k_1 cdot 0.22^k_2 cdot 0.28^k_3$
So we can write a function $f(x,y,z)=frac6!x!cdot y! cdot z! 0.5^xcdot 0.22^ycdot 0.28^z$. Since sum of all variables must be 6, we can reduce one variable. Then we want to maximaze function $g(x,y)=frac6!x!cdot y!cdot (6-x-y)!0.5^xcdot 0.22^ycdot 0.28^6-x-y$.
I don't how to do it because of the factorials. And also $k_i$ are natural numbers so i doubt that this aproach is good. I think you can examine this formula with sequences, but there is a lot of work to do then. I think there should be a theorem, which I don't know, that would help, because it was for an exam on some not mathematical (economy sth) studies and they have a small amount of math classes there.
probability
asked 2 days ago
Derm
11
I have an exercise in probability theory, which I can't solve:
There are 3 factories A B C, which produce 3 types of light bulbs. Factory A / B / C makes 40 / 20 / 40 percent of whole light bulbs. The probability of manufacturing first type of light bulb in factory A / B / C is 0.6 / 0.3 / 0.5. The probability of manufacturing second type of light bulb in factory A/ B / C is 0.3/0.4/0.2. The probability of manufacturing third type of light bulb in factory A / B / C is 0.1/0.3/0.3.
We buy 6 light bulbs. What is the most probable set of light bulbs (number of each type)? What is its probability?
So I figured out it's multinomial distribution with parameteres 6 and probabilities $p_1=0.5$, $p_2=0,22$, $p_3=0.28$, where $p_i$ is probability of buying a light bulb of $i-th$ type.
Let $X_i$ denote a number of light bulbs of $i-th$ type in the set of 6 light bulbs we bought. To solve first problem we want to maximaze $P(X_1=k_1, X_2=k_2, X_3=k_3) = frac6!k_1!cdot k_2! cdot k_3!0.5^k_1 cdot 0.22^k_2 cdot 0.28^k_3$
So we can write a function $f(x,y,z)=frac6!x!cdot y! cdot z! 0.5^xcdot 0.22^ycdot 0.28^z$. Since sum of all variables must be 6, we can reduce one variable. Then we want to maximaze function $g(x,y)=frac6!x!cdot y!cdot (6-x-y)!0.5^xcdot 0.22^ycdot 0.28^6-x-y$.
I don't how to do it because of the factorials. And also $k_i$ are natural numbers so i doubt that this aproach is good. I think you can examine this formula with sequences, but there is a lot of work to do then. I think there should be a theorem, which I don't know, that would help, because it was for an exam on some not mathematical (economy sth) studies and they have a small amount of math classes there.
probability
asked 2 days ago
Derm
11
edited 2 days ago
asked 2 days ago
Derm
11
asked 2 days ago
Derm
11
asked 2 days ago
Derm
11
11
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago
add a comment |Â
1 Answer
1
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Max $P = dfrac6!x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!.24^x_1.12^x_2.04^x_3.06^x_4.08^x_5.06^x_6.2^x_7.08^x_8.12^x_9$
with
$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$
$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$
$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$
$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$
$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$
$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$
$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$
$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$
$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$
$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i ge 0$
Ran a simulation and found out that the number and itemtypes are
$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$
Most probable with maximum probability $= 0.005971$
answered 2 days ago
Satish Ramanathan
8,84431122
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Max $P = dfrac6!x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!.24^x_1.12^x_2.04^x_3.06^x_4.08^x_5.06^x_6.2^x_7.08^x_8.12^x_9$
with
$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$
$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$
$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$
$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$
$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$
$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$
$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$
$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$
$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$
$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i ge 0$
Ran a simulation and found out that the number and itemtypes are
$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$
Most probable with maximum probability $= 0.005971$
answered 2 days ago
Satish Ramanathan
8,84431122
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
add a comment |Â
up vote
0
down vote
Max $P = dfrac6!x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!.24^x_1.12^x_2.04^x_3.06^x_4.08^x_5.06^x_6.2^x_7.08^x_8.12^x_9$
with
$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$
$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$
$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$
$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$
$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$
$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$
$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$
$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$
$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$
$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i ge 0$
Ran a simulation and found out that the number and itemtypes are
$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$
Most probable with maximum probability $= 0.005971$
answered 2 days ago
Satish Ramanathan
8,84431122
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Max $P = dfrac6!x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!.24^x_1.12^x_2.04^x_3.06^x_4.08^x_5.06^x_6.2^x_7.08^x_8.12^x_9$
with
$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$
$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$
$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$
$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$
$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$
$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$
$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$
$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$
$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$
$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i ge 0$
Ran a simulation and found out that the number and itemtypes are
$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$
Most probable with maximum probability $= 0.005971$
answered 2 days ago
Satish Ramanathan
8,84431122
Max $P = dfrac6!x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!.24^x_1.12^x_2.04^x_3.06^x_4.08^x_5.06^x_6.2^x_7.08^x_8.12^x_9$
with
$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$
$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$
$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$
$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$
$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$
$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$
$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$
$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$
$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$
$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i ge 0$
Ran a simulation and found out that the number and itemtypes are
$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$
Most probable with maximum probability $= 0.005971$
answered 2 days ago
Satish Ramanathan
8,84431122
answered 2 days ago
Satish Ramanathan
8,84431122
answered 2 days ago
Satish Ramanathan
8,84431122
answered 2 days ago
Satish Ramanathan
8,84431122
8,84431122
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
add a comment |Â
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Thank you! But this should be solved without a simulation. It was an exercise on an exam.
– Derm
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
Do you know the answer for this? Let me know how you got your probabilities? Thanks
– Satish Ramanathan
2 days ago
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7)
– Derm
yesterday
add a comment |Â
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
2 days ago
You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6times 0.5, 6 times 0.22, 6 times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$.
– BGM
2 days ago