Probability

This is a bit brute-force, but shows you what the teacher is saying... make a tree:




and carry it out 4 steps... you''ll see 16 possible outcomes. Now just count the ones that match your problem and you indeed get 5/16 and 11/16.

actually I wrote a simple program for it computing the probability by brute force.
By randomly generating 4 values - 0 for boy, 1 for girl I got the probability 5/16 and 11/16.
By randomly generating the count of girls/boys I got 2/5 and 3/5.
It doesn''t make sence to me. Why is that?

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Dave

This comes directly from "The Practice of Statistics" 1999 W. H. Freeman and Company by Daniel Yates, et al:

"If we assume the couple is equally likely to have a girl or a boy, then how unusual is it for a family to have 3 children who are all girls? If success = girl, and failure = boy, then p(success) = .5. We will define the random variable X as the number of girls. Then we want to simulate families with 3 children. Our goal is to determine the long-term relative frequency of a family with 3 girls, that is, P(X = 3)"

Then it goes on to do the simulation without using any formulas and such. But the thing is set up that you can soon be introduced to a binomial distribution. A binomial distribution is used when

"1)Each observation falls into one of just two categories, which we call "success or failure"
2) There is a fixed number n of observations.
3) The n observations are all independent. That is, knowing the result of one observation tells you nothing about the others.
4) The probability of success, call it p, is the same for each observation."

Since your situation meets this criteria, we can go ahead and use the binomial distribution. What is it? Right out of the horse''s mouth:

"If X has a binomial distribution with n observations and probability p of success on each observation, the possible values of X are 0,1,2,...,n. If k is any one of these values,

P (X = k ) =

( n! / (k!(n - k)!) ) * P^k * (1 - P)^(n-k)"

Plugging your values in for part A, where n is 4, k is 3, and P is .5:

( 4*3*2*1 / 3*2*1(1) ) * .5^3 * .5^1
= (24 / 6) * .125 * .5
= 4 * .125 * .5
= .5*.5
= .25

Thus, the probability of getting three girls is .25. To find the probability of getting a maximum of 2 girls (0,1, or 2 girls), you would simply re-use the formula and add P(X = 0), P(X = 1), and P(X = 2). Simple enough! Double check my math because I might be wrong.

If this wasn''t clear, do a google search on "Binomial setting" or "binomial distribution" or "binomial random variable". I hope this helps.

quote:

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Original post by Dave007
actually I wrote a simple program for it computing the probability by brute force.
By randomly generating 4 values - 0 for boy, 1 for girl I got the probability 5/16 and 11/16.
By randomly generating the count of girls/boys I got 2/5 and 3/5.
It doesn''t make sence to me. Why is that?

-----------
Dave

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To write a brute force program for this, generate the four random values (0 for boy, 1 for girl) and add them up. You must also define four variables to hold the frequency that so many girls show up. For example, the couple could have 0,1,2,3, or 4 girls. Maybe use an array NumGirlsFrequency[5]. Heres the psuedocode for such a program.




"Let me just ejaculate some ideas"

LuckyNewbie, thank you for your explanation. I already know about binomial distribution. Thank you for your answer.
As I mentioned in my first post, it can be computed just by counting occurences of 3 or more girls in all possibilities. I just didn''t know you have to look at the order of children.

Anyway, this probability stuff is quite strange...

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Dave