B
BodybldgDoc
How do we figure this one out?
What will be the likelihood of having exactly 1 boy in a family planning for three children?
What will be the likelihood of having exactly 1 boy in a family planning for three children?
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.
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.
(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.
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).Cool, thanks!
Ok. What if, instead of having exactly 1 boy, the family has at least 1 boy?
P(2 boys) + P(3 boys)
P(2) = 3!/2!1! * (1/2)^3 = 3/8
P(3) = 1/8
so 1/2