Probability that more girls than boys go on the field trip

[jaus tail]

Homework Statement

Homework Equations

First use equations to find out number of boys n number of girls.
Then use probability concepts.

The Attempt at a Solution

Number of boys + number of girls = 12
Boys = Girls + 2

So we get Boys = 7, girls = 5.
Select more girls than boys in a group of 3 kids so we have:
GGG, BGG, GBG, GGB
So we have: 5/12 * 4/11 * 3/10 + 3 * [5/12 * 4/11 * 7/10]

But book has done this:

Shouldn't GGG be 5 * 4 * 3 / (12 * 11 * 10)?

 

[Mark44]

Homework Statement

View attachment 222175

Homework Equations

First use equations to find out number of boys n number of girls.
Then use probability concepts.

The Attempt at a Solution

Number of boys + number of girls = 12
Boys = Girls + 2

So we get Boys = 7, girls = 5.
Select more girls than boys in a group of 3 kids so we have:
GGG, BGG, GBG, GGB
So we have: 5/12 * 4/11 * 3/10 + 3 * [5/12 * 4/11 * 7/10]

But book has done this:
View attachment 222176
Shouldn't GGG be 5 * 4 * 3 / (12 * 11 * 10)?

Your textbook is calculating the probabiity with replacement; i.e., with the events being independent. If you have a collection of 5 girls and 7 boys, the probability of choosing a girl is 5/12. If you make another selection at random, without removing the first girl, the probability of getting a girl again is still 5/12. Same for the third draw.

 

[jaus tail]

But isn't that wrong?

 

[Mark44]

But isn't that wrong?

I don't think it's wrong -- it's just a different way of doing the choosing.

 

[Ray Vickson]

But isn't that wrong?

Yes, it is wrong, if "choosing" means the usual way we employ the term! In that case all of the books possible answers are wrong, although one is numerically close.

Your solution is correct, but if I were marking it I would want to see a final answer of the form p/q and perhaps also a decimal approximation, and I would want to see a brief justification of the steps you took.
Just to be on the safe side I would want to see an explicit recognition that you are sampling without replacement, and maybe an explanation of why you are doing that.

 

[Merlin3189]

Doesn't the question specify choosing without replacement?
If a child is chosen to go on the trip, it can't be replaced and subsequently chosen again to go on the trip, else we may end up with less than 3 children.
If we persist in making more than 3 selections until we have 3 distinct children, then by ignoring the cases where a child is selected for a second (or more'th) time, we are saying that that child really wasn't replaced and available for selection.

 

[Ray Vickson]

Doesn't the question specify choosing without replacement?
If a child is chosen to go on the trip, it can't be replaced and subsequently chosen again to go on the trip, else we may end up with less than 3 children.
If we persist in making more than 3 selections until we have 3 distinct children, then by ignoring the cases where a child is selected for a second (or more'th) time, we are saying that that child really wasn't replaced and available for selection.

The question does not actually say explicitly that the choosing is done without replacement, but it does say "randomly picked", and any reasonable interpretation would be that it is done without replacement.

Anyway, you make a good point: if we can put back a previous choice (while recording that choice for future reference) we may choose the same student several times. We would need to keep picking until we got a new student not on the current list.

That brings up a nice little question: suppose we pick with replacement, but keep picking until we have three different students. What then are such probabilities as P(3 girls) or P(2 girls, 1 boy)? I am pretty sure I know the answer, and it is not too hard, but I will just leave it as an open question for now.

 

[Merlin3189]

if we can put back a previous choice (while recording that choice for future reference) we may choose the same student several times. We would need to keep picking until we got a new student not on the current list.

My point was that, if you do this, then you are not actually replacing the child in the sense used by the replacement formula..

Just because you have a formula for choice with replacement and a formula for choice without replacement, doesn't mean that either of them can be used to model this situation. You have to decide how the situation can be modeled, then, if you are so inclined, choose a formula that applies to that model.
Your description of how you would select the three children, by keeping on picking until you have the three, is not a model to which the replacement formula applies - even though you use the word replacement in your description.

 

[Ray Vickson]

My point was that, if you do this, then you are not actually replacing the child in the sense used by the replacement formula..

Just because you have a formula for choice with replacement and a formula for choice without replacement, doesn't mean that either of them can be used to model this situation. You have to decide how the situation can be modeled, then, if you are so inclined, choose a formula that applies to that model.
Your description of how you would select the three children, by keeping on picking until you have the three, is not a model to which the replacement formula applies - even though you use the word replacement in your description.

No, of course you are right. I was just agreeing with you: the concept of choosing with replacement makes no real sense in this problem, and to make it fit you would need to go to some extraordinary and artificial lengths.

However, artificial or not, extraordinary or not, my little question still stands.

 

[jaus tail]

This is horrible. I had this question in exam and in the answer sheet they gave the above solution. I wasted ten minutes in the exam trying to solve this question and ended up losing precious time. Now I got 43 out of 100 and won't get admission. Will have to apply again and study again for next year entrance exam.