Permutation of having 3 children: 2x2x2=8
There are 3 ways to have exactly one boy in 3 children.
BGG
GBG
GGB
So I am seeing 3/8.
What is the answer?
There are 3 ways to have exactly one boy in 3 children.
BGG
GBG
GGB
So I am seeing 3/8.
What is the answer?
3!/2!1! = 3 (since order doesnt matter)
1/2^3 = 1/8
So, 3/8 is the probablity of having 1 boy out of 3 tries.
1/2^3 = 1/8
So, 3/8 is the probablity of having 1 boy out of 3 tries.
Can you please explain why you did 3!/2!1!?
I can't find a good permutations website.
(3 choose 1) = # of ways to select 1 object out of 3 (without regard to order).
(n choose k) = n! / [(n-k)!*k!]
So n! would represent the # of ways to select all 3 objects in any order. When you divide by (n-k)! you are cutting it off after k objects. So instead of considering n objects to select, you consider the first k objects.
Ex: 6 choose 2. The n! = 6! means you want to select all 6 objects. But the (n-k)! = (6-2)! = 4! means you only consider the first k = 2 objects: 6!/4! = 6*5. There are 6 choices for the first selection and 5 for the second selection. Then you stop.
The 1/k! accounts for the # of ways to order the objects. With k objects you can order them k! ways (there are k ways to choose which comes first, k-1 ways to choose which comes second, etc). So if you don't care about order, you divide by k! because that's how many times each arrangement is 'duplicated' in your selection process, so to speak.
Ex: If you select 3 objects from 10, you could pick A B C or B C A or B A C to name a few. But those are the same 3 objects. Since there are 6 ways to arrange the letters A, B, and C, you divide by 6 (3!) to account for this. Only 1 in every 6 selections will be a different group of 3.
So what if the OP cared about order? Could you please show me by incorporating it into his question? Thanks. Mr. 28 QR.
You just don't divide by k!. Instead of n!/[(n-k)!*k!] you have n!/(n-k)!
So it would be 3!/2! = 3 still... and it's the same because you are only looking for ways to select one kid to be a boy out of the three... and there is only one way to order one object.
If you needed something like 5 items out of 12 with and without order mattering, then you'd have:
(12 choose 5) = 12!/7!5! = 792 ways to choose without regard to order.
And then 12!/7! = 95,040 ways to choose if order matters.
In this case you are selecting 5 objects. If order matters then you could choose A, B, C, D, E or A, C, D, E, B or D, B, A, E, C etc. How many ways can you arrange 5 objects? You can pick any of the 5 to go first, then 4 to go second, 3 to go third, 2 to go fourth, and 1 to go last. That's 5! = 120 ways. So I can select objects A B C D and E in 120 different ways. If order matters then that's fine - we want to consider those separate ways of choosing them. But if order doesn't matter, we only need to consider 1 out of those 120 ways. That's why you divide 95,040 by 120 (or 5!) and get 792.
Cool, thanks!
Ok. What if, instead of having exactly 1 boy, the family has at least 1 boy?
Ok. What if, instead of having exactly 1 boy, the family has at least 1 boy?
7/8.
At least 1 boy means, not getting all three girls.
There are 8 possibilities of "making" children but only 1 way to make all 3 girls.
8-1 gives you 7, so it is 7 out of 8.
At least 1 boy means, not getting all three girls.
There are 8 possibilities of "making" children but only 1 way to make all 3 girls.
8-1 gives you 7, so it is 7 out of 8.
Advertisement - Members don't see this ad
Then that's the probability of not having 0 boys. The probability of having 0 boys is (3 choose 0) * (1/2)^3 or 1/8. Then since you want the probability of this not happening, it's 1 - (1/8) = (7/8).
So then what would it be to have at least 2 boys out of 3 tries?
Thanks 🙂
Thanks 🙂
Yeah since P(2 or 3) = P(2) + P(3) because they are independent events.
the binomial expension I use for most of these "kids" problem
since you can only have a boy or a girl.. the probability of having a girl is always 1/2. the probability of having a boy is also always 1/2
a= girl; b=boy
example
2 girl 3 boy total 5 kids
(a+b)^5= (a^5)+5(a^4b)+10(a^3b^2)+10(a^2b^3)+5(ab^4)+b^5
so you would select 10(a^2b^3) and replace both (a) and (b) with 1/2
..... Hence for a set of kids with 1 being a boy then 2 must be girls you would expend
(a+b)^3= (a^3)+3(a^2b)+3(ab^2)+(b^3)
you would use this part of the expension: 3(a^2b)
and replacing (a) and (b) with 1/2 you would end up with : 3/8 as athe probablity of having 1 boy and 2 girls
or you could use P=(n!/s!t!)a^s*b^t
where (s) is the number of girl and (t) the number of girl and 👎 the total number of kids
which would also give you 3/8
BUT
when it comes to at "least x kid" problems these equations seem to be useless.. so what are your take on this matter Streetwolf? are these genetics equations really useless?
since you can only have a boy or a girl.. the probability of having a girl is always 1/2. the probability of having a boy is also always 1/2
a= girl; b=boy
example
2 girl 3 boy total 5 kids
(a+b)^5= (a^5)+5(a^4b)+10(a^3b^2)+10(a^2b^3)+5(ab^4)+b^5
so you would select 10(a^2b^3) and replace both (a) and (b) with 1/2
..... Hence for a set of kids with 1 being a boy then 2 must be girls you would expend
(a+b)^3= (a^3)+3(a^2b)+3(ab^2)+(b^3)
you would use this part of the expension: 3(a^2b)
and replacing (a) and (b) with 1/2 you would end up with : 3/8 as athe probablity of having 1 boy and 2 girls
or you could use P=(n!/s!t!)a^s*b^t
where (s) is the number of girl and (t) the number of girl and 👎 the total number of kids
which would also give you 3/8
BUT
when it comes to at "least x kid" problems these equations seem to be useless.. so what are your take on this matter Streetwolf? are these genetics equations really useless?
