Please help me with this math (NSR)

[alto]

Suppose, for the purposes of the following question, that each child born to a couple is equally likely to be a boy as to be a girl regardless of the sex distribution of the other children in the family.


If a couple has 5 children what is the probability that exactly three of the children are boys?

and if a couple has 5 children what is the probability that there is at least one girl?


These should be easy, I for some reason cannot figure it out. Anyone?

[Telephil]

542

[Viva]

Read & learn:

http://www.mathgoodies.com/lessons/v...obability.html

[Dingleberry]

The answer is always 47.

[grrrr]

Pascal's triangle

1
1 1=50%b, 50%g
1 2 1=25%bb, 50%bg, 25%gg
1 3 3 1, etc
1 4 6 4 1 etc
1 5 1 0 10 5 1 etc

10/32 exactly gggbb
1/32 only bbbbb

However, if you need to use formulas, you're on your own. I have only so much time to waste right now.

[The AD]

Originally posted by alto
Suppose, for the purposes of the following question, that each child born to a couple is equally likely to be a boy as to be a girl regardless of the sex distribution of the other children in the family.


If a couple has 5 children what is the probability that exactly three of the children are boys?

and if a couple has 5 children what is the probability that there is at least one girl?


These should be easy, I for some reason cannot figure it out. Anyone?

Maybe I'm missing something here, but I think there are only 6 possible combinations: 5B, 4B/1G, 3B/2G, 2B/3G, 1B/4G, 5G.

So there is exactly a 1/6 chance of having three boys and two girls, and a 5/6 chance that there will be a t least one girl.

The math should be easy, but I can't remember. I think it involves factorials, though!

[random]

The way AD did it is the way I learned last year. I think. But I spent much of my stats class sleeping...

[grrrr]

Originally posted by The AD
Maybe I'm missing something here, but I think there are only 6 possible combinations: 5B, 4B/1G, 3B/2G, 2B/3G, 1B/4G, 5G.

So there is exactly a 1/6 chance of having three boys and two girls, and a 5/6 chance that there will be a t least one girl.

The math should be easy, but I can't remember. I think it involves factorials, though!

Yes, you are missing something.

Yes, it can be done with factorals, however the Pascal triangle gives the same answer.

Think about it. There 10 possible ways of having three girls GGGBB, GGBBG, GBBGG, BBGGG, BGBGG, BGGBG, BGGGB, GBGBG, GGBGB, GBGGB , but only one way of having all boys BBBBB. All of those possibilities have equal chances of occurring, therefore it must be ten times as likely to have three girls as all boys.

[dipstik]

Did you guys know that if you take the straight length of a river ( the "as the crow flies" distance), and divide it by the true length (the length you would travel in a raft), you get Pi?

[Mountain Junkie]

Originally posted by alto

If a couple has 5 children what is the probability that exactly three of the children are boys?

Sample Set is 2^5=32. Number of ways to choose 3 boys is 5C3, or 10. So, 10/32.

and if a couple has 5 children what is the probability that there is at least one girl?

Probability that there are NO girls in family: 1/32, so at least one girl is 1 - 1/32, or 31/32.

[The AD]

Originally posted by grrrr
Yes, you are missing something.

Yes, it can be done with factorals, however the Pascal triangle gives the same answer.

Think about it. There 10 possible ways of having three girls GGGBB, GGBBG, GBBGG, BBGGG, BGBGG, BGGBG, BGGGB, GBGBG, GGBGB, GBGGB , but only one way of having all boys BBBBB. All of those possibilities have equal chances of occurring, therefore it must be ten times as likely to have three girls as all boys.

I don't think so. You're assuming it's order dependent, and it isn't. The question wasn't "what's the probability of having three boys in a row?" I see only 6 possible outcomes, with exactly one of them being 3 boys and 2 girls.

EDIT: OK, I think I get it now. I guess I better go back to school! I should have known I was wrong because I couldn't explain it mathematically.

[bigAK]

The AD's way is wrong (sorry), doesn't account for the fact that not all cases happen with equal probability. here's the way i would do it:

number of total samples = number of possible outcomes^number of trials = 2^5 = 32

now, for exactly three boys, the possible samples are:
bbbgg, bbgbg, bbggb, bgbbg, bgbgb, bggbb, gbgbb, gbbgb, gbbbg, ggbbb.
the number of samples = 10.
probability = number samples/total number samples = 10/32

for at least one girl (P), it is easier to consider the case of no girls (with probability x) and P=1-x.
x = number of samples with no girls (only bbbbb)/total number of samples = 1/32.
therefore P = 1- (1/32) = 31/32

grrr's pascal method is the shortcut version of this, but if you don't know it, it may be hard to understand. this way just uses basic definitions of probability theory...

edit, although i can see i'm way too slow at typing...

[Ireallyliketoski2]

I need some help with this.

Say you have two kids, what's the probability that there is exactly one boy, with a girl and a boy being the only options. You have 4 possible out comes
bb
bg
gb
gg
So the probability of having exactly one boy = 2/4 = 1/2

If you use combinations on the above example you use 2C1, correct?

which would give you

2!/(1!(2-1!)) = 2,

which makes sense.

But when I extend the problem to consider that a family can have 3 different types of kids, say boy, girl, or tiger. Same question what is the probability that there is exactly 1 boy.
The sample is now 3^2=9 and we have

bb
bg
gb
gg
tg
tb
bt
gt
tt

So we have our probability of having exactly one boy if 4/9. Where I'm confused now is what combination would apply to this problem. This choose that, but what choose what. In the previous problem we were saying 5 choose 3, is that because there were 5 kids and we wanted 3 boys? But that wouldn't work here because then we would still be doing 2 choose 1. I always struggle with determing this and whether to use permutations or combinations. Any help to set me straight would be much apprecitated?

[frozenwater]

wow. I don't know what the hell you guys are talking about. So I will get out my degree and start polishing it. This reminds me that I am smart.

[Buzzworthy]

Word Problems..........nuff said.

[bigAK]

for the problem (ireallyliketoski's) above:

the b,g combination actually includes a "hidden" factor of one, representing the possibilities of the other child (only 1, g). the problem is actually ((2C1)*1)/4. there are 2C1 possibilities for exactly one boy, and 1 possibility for the other child. you have to "fill" the two spots in the family, and there is only 1 way to do so maintaining the constraints of exactly one boy.

now, for your example with b,g,t, there are again 2C1 possiblities for exactly one boy. but now there are two possibilitites of what the other child is (in other words, you can "fill" the family with 2 possible ways). so the problem looks like ((2C1)*2)/9

consider poker hands. to find the possibility of exactly one pair:
first, choose the pair. there are 13 cards and 4 suits, 2 cards make a pair. so the possibilities are 13*(4C2). but, you're not done because you also have to select 3 more cards to "fill" your hand. 12C3 (3 denominations of other cards out of 12 possible) and 4^3 ways of picking the suits (4 ways for the first "other" card, 4 ways for the second, etc.) so, the total number of ways of picking exactly one pair:
13*4C2*12C3*4^3
total number of ways of picking 5 cards from 52 is 52C5, so the probability of exactly one pair is
(13*4C2*12C3*4^3)/(52C5)

hopefully this makes some sense...

[britney]

heres how i did it (there must be an easy way but i couldnt think of it):

1.Probability of exactly 3 boys-- not order dependent. An event of 3 boys is exactly the same as an event of exactly 2 girls. The
first girl (Sarah) can be in any one of 5 positions and then the second girl (Amy) can be in
any one of the remaining 4 positions, or there are 5*4 = 20 possible permutations, BUT
this double-counts the possibilities because for our purposes, Sarah, Amy, boy, boy, boy
is the same as Amy, Sarah, boy, boy, boy. So the actual number of ways for there to be
exactly 2 girls is 5*4/2 = 10. So now we have the answer: There are 10 ways for 5
children to be 2 girls and 3 boys, and each way has probability (0.5)^5. the probability of
exactly 2 girls is 0.313, or 31.3%.

2.Probabiliyt of at least one girl--The event "at least one girl" is the opposite of the event "all
boys"-- it's easier to figure the probability of all boys, which is (0.5)^5. So the probability
of "at least one girl" is 1-(probability of all boys) = 1-0.5^5 = 1- .0313 = .969 or 96.9%

3. If the family has 5 children, the probabliity that the parents are utarded is the same as #2 above, or 96.9%



-b

[grrrr]

Originally posted by britney
The
first girl (Sarah) can be in any one of 5 positions and then the second girl (Amy) can be in
any one of the remaining 4 positions

-b

Wow. I like how you do math.

[swiss powda]

Originally posted by dipstik
Did you guys know that if you take the straight length of a river ( the "as the crow flies" distance), and divide it by the true length (the length you would travel in a raft), you get Pi?

not true, or at least not true as you stated it. length of straight(as the crow flies) river would be shorter than true (raft) length giving you a number less than 1, you might get 1/pi, but I doubt it seeing as some rivers are curvier than others.

Just my common (2) sense.

[DJSapp]

EDIT: I'm a retard, and don't know what I was thinking last night. This is wrong. Whiskey + proabaility = 47

Ummmm, actually this problem is really easy.

Order is not important, just the final outcome.

6 possible situations, GIVEN THERE ARE 5 CHILDREN, WITH EQUAL PROABILITY OF EITHER SEX.

5 boys
4 boys, 1 girl
3 boys, 2 girls
2 boys, 3 girls
1 boy, 4 girls
5 girls

Proability that there are exactly 3 boys, 1/6

Proability that there is one or more girls,
5/6

Do I get a gold star?

[bigAK]

Originally posted by DJSapp
Do I get a gold star?

sorry no gold star, read the posts above...

[KrisSkis]

I hate Math...damnit!!!

[grrrr]

There have been three correct ways to do this listed, and we still have some people seeing 6's. There is not an equal chance of BBBBB and BBGGG:

Chance of the first child being a boy: 50%
Chance of second child being a boy if the first was a boy: 50%x50% = 25%
Chance of third child being a boy after two boys: 50% x 25% = 12.5%
Chance of 4th: 50% x 12.5% = 6.25%
Chance of 5th: 50% x 6.25% =3.125% or 96.875% (or 31/32) of at least one girl.

[Ireallyliketoski2]

Yes, that makes total sense, thanks AK. I always loved probability, in college I had a probability class and my professor was a gambling freak. We studied the Roulette wheel and determined that the expected value for every bet on the game (black, red, odd, even, single number, etc.) has an expected value of -0.0526, that means that mathematically the casino gets a nickel of every dollar bet on the table. Maybe that's why I had problem with the boy girl problem, thanks to my teacher I can only figure out gambling problems.