Probability of Girl in Family of 3 Children

AI Thread Summary
In a family with three children where the eldest is a boy and at least one other boy exists, the probability of having at least one girl among the children is calculated to be 2/3. The possible combinations of children are BBB, BBG, and BGB, excluding BGG since it contradicts the given information. Each of these combinations is equally likely, and two out of the three combinations include at least one girl. Therefore, the initial assumption of a 0.5 probability is incorrect; the correct probability is 2/3. This conclusion emphasizes the importance of considering all possible combinations when calculating probabilities.
jackbauer
Messages
10
Reaction score
0
Hi people.
I have a short question from a homework. given a family with 3 children, the sexes of which are unknown: if the eldest child is a boy and there is at least one other boy, then what is the probability one child is a girl?
To me it seems obvious that there are at least two boys from this info, then either there are 3 boys or 2 boys and 1 girl, so the probability one child is a girl is 0.5. Is this method correct?
Thanks,
JB
 
Physics news on Phys.org
A family with three children and the eldest is a boy. Writing B for boy, G for girl, from oldest to youngest, the possibilities are:
BBB
BBG
BGB
BGG
Assuming "boy" or "girl" are equally likely at each birth, these are equally likely.

But we are told that at least one child is a boy: that throws out BGG leaving
BBB
BBG
BGB and they are still equally likely.
Since 2 of those 3 correspond to "one child is a girl", the probability that one of the children is a girl is 2/3, not 1/2. Yes, it is true that there must be "either 3 boys or 2 boys and a girl", but those are NOT "equally likely".
 
As far as I'm concerned you should better explain the underlying reason: probability of each sex child birth is fifty-fifty. In a formal proof is often request.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

Chance of Selecting Twins from Group of 30 [Solved!]
Replies
24
Views
2K
Probability - family of two children
Replies
4
Views
4K
I Probability of 2 Different Gendered Children Born in Order
Replies
10
Views
4K
Solving the Boy Girl Paradox Probability for Random Families
Replies
9
Views
4K
Probability of At Least 1 Girl Child in Family of 3
Replies
11
Views
3K
Conditional Probability: probability of having a girl.
Replies
24
Views
5K
What Is the Probability That Exactly 3 Out of 8 Children Are Girls?
Replies
2
Views
2K
I What Is the Counterintuitive Probability in the Bag of Balls Problem?
Replies
18
Views
3K
What is the Probability of Gender Combinations in a Family with Four Children?
Replies
2
Views
3K
I Does the statistical weight of data depend on the generating process?
4
Replies
174
Views
11K

Hot Threads

Recent Insights

Back
Top