The Official Math/Physics/Whatever Homework questions thread

[bigpooch]

Originally Posted by furyshade

quick set theory question, is U{{a,b,c,d,e,f},{e,f}} equal to {a,b,c,d,e,f} or {a,b,c,d,e,f,e,f}?

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[furyshade]

Originally Posted by bigpooch

.

that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?

[bigpooch]

Originally Posted by furyshade

that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?

Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc.

On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}.

There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.

[furyshade]

Originally Posted by bigpooch

Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc.

On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}.

There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.

so would the the set of natural numbers be a transitive set?

[bigpooch]

Originally Posted by furyshade

so would the the set of natural numbers be a transitive set?

If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.

[furyshade]

Originally Posted by bigpooch

If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.

alright, that makes sense. the questions says to show that {null,{null}} is transitive and then asks for an example of an infinitely large transitive set so i figured that is what they were getting at. thanks a lot for the help!

[thylacine]

Originally Posted by bigpooch

If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.

Are you sure?

[smcdonn2]

If probability of having a boy is 1/2. then what is the probability of having three children of the same sex?

I think it is:

1-2[(4 choose 1)(1/2)^1(1/2)^3]

I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy

Is this right?

[smcdonn2]

Lost on this one, Just a move in the right direction would be great

Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively.

Theorem 3.2.1 Binomial Distribution

(n choose k)(p)^k(1-p)^n-k

[Wyman]

Originally Posted by smcdonn2

If probability of having a boy is 1/2. then what is the probability of having three children of the same sex?

I think it is:

1-2[(4 choose 1)(1/2)^1(1/2)^3]

I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy

Is this right?

Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other?

If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2.

Which is also what you got, but idk why you subtracted from 1.

[Wyman]

Originally Posted by smcdonn2

Lost on this one, Just a move in the right direction would be great

Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively.

Theorem 3.2.1 Binomial Distribution

(n choose k)(p)^k(1-p)^n-k

You've got N trials, k_1 of which need outcome 1: N choose k_1
Now you've got N-k_1 remaining, k_2 of which need outcome 2: (N-k_1) choose k_2

The probability of getting k_1 outcome 1's, then k_2 outcome 2's, then outcome 3's is: (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)

So your probability is:
(N choose k_1) * ((N-k_1) choose k_2) * (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)

[smcdonn2]

Originally Posted by Wyman

Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other?

If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2.

Which is also what you got, but idk why you subtracted from 1.

Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.

[smcdonn2]

a college awards 5 scholarships
there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award?

I did it like this but got a difference answer than the book

(8C0)(10C5)/(18C5)=2.94

1-2.94=97.06%

the book says 96.4

[Wyman]

Originally Posted by smcdonn2

Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.

This makes no sense, and you got lucky because 1/2 = 1-1/2.

[Wyman]

Originally Posted by smcdonn2

a college awards 5 scholarships
there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award?

I did it like this but got a difference answer than the book

(8C0)(10C5)/(18C5)=2.94

1-2.94=97.06%

the book says 96.4

p(all men) = 8C5 / 18C5

p(all women) = 10C5 / 18C5

p(both men and women) = 1 - p(all men) - p(all women)