Probability of At Least 1 Girl Child in Family of 3

[1/2"]

Homework Statement

"In a family of 3 children , what is the probability that they will have at least 1 girl child?"
This is the problem.Simple enough.But i am having a problem in finding out the total no. of possibilities.

Homework Equations

The Attempt at a Solution

I got the total possibilities as {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}
(here b is for boy and g for girl ) ie i am getting total 8 possibilities
probability=7/8
But my teacher says it should be 3/4 because possilbilities will be
{3b,3g,2b and 1g,2g and 1b }
I am totally confused.I can't reason out which should be the correct one.
Please help.

 

[Simon Bridge]

bbb = (3,0)
bbg = (2,1)
bgb = (2,1)
bgg = (1,2)
gbb = (2,1)
gbg = (1,2)
ggb = (1,2)
ggg = (0,3)

P(G=0)=1/8
P(G=1)=3/8
P(G=2)=3/8
P(G=3)=1/8

but that's where order matters - this counts the situation where the girl is born second is different for when the girl is born first. But the question counts both of these as one girl. So your possible combinations are actually:

bbb = (3,0)
bbg = (2,1)
bgg = (1,2)
ggg = (0,3)

so P(G=3)=1/4, P(G=2)=1/4, P(G=1)=1/4 for a total P(G>0)=3/4.

 

[ehild]

It matters what events are equally probable. Usually boys and girls are born with equal probability. So the family has equal probability to have either a boy or girl at every birth, 8 equally possible outcomes. The order counts. The OP is right.

ehild

 

[Curious3141]

Yes, the OP is right here.

 

[Ray Vickson]

Homework Statement

"In a family of 3 children , what is the probability that they will have at least 1 girl child?"
This is the problem.Simple enough.But i am having a problem in finding out the total no. of possibilities.

Homework Equations

The Attempt at a Solution

I got the total possibilities as {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}
(here b is for boy and g for girl ) ie i am getting total 8 possibilities
probability=7/8
But my teacher says it should be 3/4 because possilbilities will be
{3b,3g,2b and 1g,2g and 1b }
I am totally confused.I can't reason out which should be the correct one.
Please help.

This problem is the same as asking for the probability of getting at least one 'head' in 3 tosses of a fair coin. The relevant distribution is the so-called Binomial Distribution (eg., http://stattrek.com/lesson2/binomial.aspx ). That shows that your answer of 7/8 is correct.

RGV

 

[Simon Bridge]

there's more different ways to get particular combinations - OP and teacher were thinking of different situations. Maybe the question was not written as carefully as it should have been?

 

[cepheid]

there's more different ways to get particular combinations - OP and teacher were thinking of different situations. Maybe the question was not written as carefully as it should have been?

I'm not sure I understand your argument. Assuming the family is having 3 children sequentially (one after the other), then bgb and gbb are two distinct outcomes, and both contribute to the set of possible outcomes in which you have exactly one girl. So I agree with the original poster (OP), ehild, Curious3141, and Ray Vickson about how to do the counting. The coin toss analogy is what clinched it for me.

 

[Ray Vickson]

there's more different ways to get particular combinations - OP and teacher were thinking of different situations. Maybe the question was not written as carefully as it should have been?

Since Pr{>= 1 girl} + Pr{0 girls} = 1, the teacher would need to have P{0 girls} = 1/4 in a 3-child problem. In fact, Pr{0 girls} = 1/8.

RGV

 

[awkward]

bbb = (3,0)
bbg = (2,1)
bgb = (2,1)
bgg = (1,2)
gbb = (2,1)
gbg = (1,2)
ggb = (1,2)
ggg = (0,3)

P(G=0)=1/8
P(G=1)=3/8
P(G=2)=3/8
P(G=3)=1/8

but that's where order matters - this counts the situation where the girl is born second is different for when the girl is born first. But the question counts both of these as one girl. So your possible combinations are actually:

bbb = (3,0)
bbg = (2,1)
bgg = (1,2)
ggg = (0,3)

so P(G=3)=1/4, P(G=2)=1/4, P(G=1)=1/4 for a total P(G>0)=3/4.

Simon Bridge,

In the second part (P(G>0) = 3/4), you are making the mistake of assuming that all the combinations (3,0), (2,1), (1,2) and (0,3) are equally likely. They are not, as your analysis in the first part shows.

 

[Simon Bridge]

Yeah I've acknowledged that in post #6.
Apologies if that was unclear.
I'm not actually disagreeing with you - I'm trying to understand why the teacher would insist on this particular answer.

maybe the teacher is thinking of a different kind of problem, one where the order does not matter. eg. what if the kids were triplets? What if there were all adopted at the same time? Maybe the teacher made a mistake? Maybe the question was poorly worded? Maybe the question was misread or out of context?

I suspect a confusion with combination/permutation counting.
We are only told that the family has three kids, not how they got there.
And I think that's the source of the confusion.

If you had a triple-cone ice-cream but only two flavors - and you don't care the order they are applied, then this would hold. There are four possible combinations and you get one of them.

But: if you put three scoops on the cone at random and ask the probability you got at least one chocolate, then there are 7 ways to get that out of eight possible permutations.

OR maybe the teacher is thinking about the boy-girl paradox (which involves only 2 siblings) instead.

 

[awkward]

The difference between the "sex of children" problem and the ice cream flavors problem is that with the children we assume that their sexes are independent. If we also assume that boys and girls are equally likely, then we end up with P(at least one girl) = 7/8.

For the ice cream flavors problem, the assumption you seem to have in mind is that each combination of flavors is equally likely. That's a reasonable assumption, but it's different than assuming, for example, that the flavor of each scoop is chosen independently of the others and all the flavors are equally likely.

 

[ehild]

Probability is defined for an experiment with equally probable outcomes. These are the "elementary events" and all other events are some combinations of them. The probability of an event is the number of favourable combinations of the elementary events divided by the number of all possible outcomes. Throwing a dice has 6 equally probable outcomes, getting a number greater then 4 can happen in two ways (5 OR 6) so its probability is 2/6.

Let be the experiment that a person asks three scoops of ice-cream out of two flavours. All scoops of the same flavour are identical. The consumer might say "two bananas and one vanilla": . If the order does not count, (you get the ice cream on a plate) the possible outcomes of equal probability can be bbb, bbv, bvv, vvv. Giving the ice cream in a scone, the consumer might want a special order. He wants to start and finish with banana, for example. in this case, there are 2^3 equally possible outcomes.
In case of families having boys and girls, the experiment might be that the couple decides to have three babies. The basic experiment is having a baby, and it has two outcomes, boy or girl, if we neglect the possibility of twins. (But even twins do not come to the world at the same instant, unless Siamese twins.) The parents repeat the experiment three times. The possible outcomes are bbb, bbg, bgb, gbb, bgg, gbg, ggb,ggg. Having at least one girl means all possibilities except bbb: 7 altogether. Having at least one girl is the same as not having three boys. The probabiliti is 7/8.

The teacher would be right in case of the following experiment: we invite four people. One has three boys, the other three girls the third has two boys and a girl, and the fourth has two girls, one boy. Somebody, not knowing about the children, picks one person in random. What is the probability that he chooses a person with at least one girl?

ehild