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 Maths Question....PROBABILITY DISTRIBUTIONS
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| Author | Topic: Maths Question....PROBABILITY DISTRIBUTIONS | ||
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Pro Bodybuilder 
Posts: 378 |
can someone help me finish my assignments , this question is pretty basic if you know the formula, i just cant find my notes from earlier on in the year, 1/ What is the probability of a 5 child family being 3 boys and two girls? is it like a xCx question, i cat quite remember the formula, using the n! button??     | ||
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Amateur Bodybuilder 
Posts: 272 |
This link might help you out. http://library.thinkquest.org/10030/4pcpac.htm Good luck. -Warik | ||
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Cool Novice 
Posts: 19 |
It's been a while since I've done discrete, so I don't have any formulas handy in memory, but I'm *pretty* sure you can look at the problem this way: What are all the different possible combinations of boy/girl familes of 5 children? Boys Girls So take the combination of 3 boys and 2 girls out of the total and you have a 1/6 probability. I'm not 100% certain that that is the correct way to tackle that problem, but I think it is. Do me a favor and let me know when you find the answer. | ||
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Amateur Bodybuilder
Posts: 231 |
Phoenix, Beautifully done. May1010 | ||
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Pro Bodybuilder
Posts: 538 |
It depends... 2 ways to answer: Order doesn't matter: 1 in 6 mmmmm order does matter: 10 out of 32 or 31.25% mmmmm mmmmf mmmff fffmm ffffm fffff SV | ||
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Pro Bodybuilder
Posts: 538 |
Now I think of it, I believe it's the latter.. 31.25% looks right.. You cant just look at the combinations, you have to look at the permutations not just the combinations. | ||
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Elite Bodybuilder 
Posts: 988 |
it is 1 in 6. | ||
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Pro Bodybuilder
Posts: 538 |
Sorry MP5, no it isnt.... | ||
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Cool Novice
Posts: 25 |
I see a flaw here, for it to be 1 in 6, it would have to be equally probable that it could be any combination of m/f as all one gender.. given that each time it's 50/50 chance that logic doesn't seem right to me. I am no mathematician, that just seems right to me. peace, ILLmatic | ||
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Guru 
Posts: 4190 |
The equation you reference cannot be that simple. The probability of a boy vs a girl in birth is not readily predictable. Since it is the father that determines the sex, it will be dependent upon the father's ratio of x-chromosome sperm to y-chromosome sperm, in addition it is also dependent upon the strength of the sperm, given that y-chromosome sperm are stronger than the x-chromosome sperm. Then there is the resistance of the femal egg to either x or y chromsome sperm. So, to predict what the liklihood of 3 boys and two girls is not easily predictable even among relatives. For example, my first wife we had three girls, with her second husband (freak) she had two more girls. She always wanted a boy and failed. Now my second wife completely different genetics she had two boys. So, with the same father there are three girls and two boys (but different mothers) My uncle had 4 girls and 1 boy, my other uncle had 1 boy and 1 girl, and I have 1 brother and 1 sister. The closest statistical number could only really be derived by a comparison of the actual number of families with the required combination of children vs the rest of the population with five children, but the wrong combination. And even this will not give you a tremendously reliable number. | ||
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Cool Novice
Posts: 19 |
I see your point Chesty. I think the question that needs to be asked is if this assignment is for a genetics course or for a mathematics course. If it's a genetics course then we're going to need additional information about the mother and father in order to determine the probability. If the question is from a math course, then I imagine it's pretty safe to say that we can assume a 50% probability of the child being a boy each time. | ||
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Guru
Posts: 4190 |
True and the answer is purely a simple statistical one | ||
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Amateur Bodybuilder
Posts: 253 |
it's been a while since i took probability, and i can't type some of the symbols, but here goes. i believe it's a binomial distribution and the odds are (assuming 50/50 chance for boy/girl): 0 boys: 1/32 essentially the odds on n boys are "5 choose n" times p to the n power times (1-p) to the 5-n power. in this case p and 1-p are both .5
so here we have 5!/3!2! * (.5)^5 or 10/32 | ||
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Pro Bodybuilder
Posts: 538 |
Runner.. If you read up that's exactly what I had said.. I even took the time to write out all the possibilities.. it's 10/32 or 31.25% end of discussion | ||
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Pro Bodybuilder
Posts: 378 |
thanks to all that tackled the question, chesty it is a purely statistical one, thanks runner i think your on the right track, i remember when we first tackled this all those months ago it was using the factorials on the calculator. | ||
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Pro Bodybuilder
Posts: 378 |
thanks to you steroid virgin, im just having trouble with the calculation so the answer is: out of the 5 children what are the Chances (c) of 3 boys | ||
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Pro Bodybuilder
Posts: 330 |
ACTUALLY, THATS A TERRIBLE QUESTION FOR A STATS CLASS... YOU "HAVE" TO GO WITH CHESTY ON THIS... EACH PERSON DOES NOT HAVE A 50-50 SHOT AT POPPING OUT A MALE OR FEMALE... GENETICS WILL PREDETERMIN THE PROBABILITY OF THE CHILDS SEX... I THINKN YOU SHOULD JUST GET FUNKY ON YOUR MATH TEACHER AND PISS HIM OFF... BUT HAVE THE PURE STATISTICAL ANSWER HANDY AS WELL, BUT DEFINITELY SAY THAT THE QUESTION IS TWO SIDED WITH MANY MANY POSSIBLE ANSWERS... SO YOU GET IT RIGHT AND YOU PISS HIM OFF... WHATS BETTER THEN THAT? | ||
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Cool Novice
Posts: 19 |
i need to change my answer as runner is correct; it is a binomimial distribution. the complete equation is: (N!/(x! * (N-x)!)) * (p^x) * ((1-p)^(N-x)) where N is the total number of tries (5), x is the number of successes (in this case boys, so 3), and p is the probability of success (.5). barney, if you check, runner plugs in the number and solves the equation at the end of his post. | ||
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Amateur Bodybuilder
Posts: 233 |
I feel so stupid now. | ||
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Cool Novice
Posts: 19 |
quote: kat, a while back you responded to a philosophy thread and posted about nietzsche, so i imagine it's safe to say that you're educated. if you didn't major in math or an engineering field, then chances are you just haven't been exposed to much probability. | ||
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Pro Bodybuilder
Posts: 538 |
You are doing nothing wrong, you just are not finishing... the problem is two part! First you determine the number of successful outcomes (permutations) 5Chance3, which you had 10, thats right.. ((5!)/3!*2!)= 60/6 = 10 Then the second part is determining the number of all possible outcomes.. Which is simply 2^5 (2 to the 5th power) = 32. the divide the successful outcomes (10) by the total possible (32), you get 10/32, or 31.25% Thanks
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Amateur Bodybuilder
Posts: 233 |
I always thought probability was like Fate...guess they came up with an equation for it...and you're right!!! I got a C in first year college business math and busted my nuts for that...I am a complete loser at math. Which I understood trig though...could have gotten first place in my billiards tourney instead of third...  | ||
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Pro Bodybuilder
Posts: 538 |
If you were a complete loser, you would have failed business math.. dont be so tough on yourself.. Everyone has their own skills and talents kat. | ||
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Pro Bodybuilder
Posts: 538 |
bump for the guy who posted the question in the first place.. |
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posted December 06, 2000 09:05 AM