Probability questions- plz help

[Xiang]

If the probability of having a male child is 0.5 and the probability of having a female child is also .
suppose a family stops when there are 2 male children or when they have 4 kids in  all.
what is the sex ratio of a population with such a family planning?

thanks :)

[Xiang]

probability of female child is also 0.5

[Akshay Jain]

sample space is - [(BB)(GBB)(GGGB)(GGGG)(GGBB)(BGGB)(BGB)(GBGG)]
count the total no of grls n boys in sample space
boys-12 n grls- 16
ratio 3/4
(m nt sure bout it)

[Maahi]

M getting 1

[Sinistral]

there cant be more than 2 male children. i.e. either 0 male child or 1 male child or 2 male children.

2 kids family:
possible only if: MM

3 kids family:
Possible only if last child is a male and either of first 2 are males ie FMM, MFM

4 kids family:
0 male child:  FFFF
1 male child:  MFFF,FMFF,FFMF,FFFM
2 male children:  (last child is male) : MFFM,FMFM,FFMM
 
Ms=16
Fs=24
sex ratio = no females per 1000 male = (24/16)*1000=1500

[Xiang]

@akshay- what about the combinations- GBGB, BGGG?

[Xiang]

The answer is equal number of girls and boys.

:/ how?

[Amit Goyal]

If the probability of having a male child is 0.5 and the probability of having a female child is also .
suppose a family stops when there are 2 male children or when they have 4 kids in  all.
what is the sex ratio of a population with such a family planning?

We can do it in multiple ways:
One is by quick observation. First child is equally likely to be a boy or a girl. So, half of the first borns' are boys and half for them are girls. Second child is again equally likely to be a boy or a girl. Again half of the second borns' are boys and half for them are girls. Now  only families with fewer than two boys so far will have the third child. However, even in this case, since the sex of the third child is independent of the first two, we have half of the third borns' are boys and half of them are girls. And similarly for the forth. Thus  1/2 of the population is male and half is female. Thus, the sex ratio is 1.
 
Alternatively, we can also do it by sample space method. Check that the sample space is given by,
S= {BB, BGB, GBB, GGBB, GBGB, BGGB, BGGG, GBGG, GGBG, GGGB, GGGG}
and the associated probabilities are:
Pr(BB) = 1/4
Pr(BGB) = Pr(GBB) = 1/8
Pr(GGBB) = Pr(GBGB) = Pr(BGGB) = 1/16
Pr(BGGG) = Pr(GBGG) = Pr(GGBG) = Pr(GGGB) = Pr(GGGG) = 1/16
Probability can also be interpreted as proportion of families. For example, Pr(BGB) is the probability that the family will have first boy child, second girl child and third boy child or the proportion of families with first boy child, second girl child and third boy child.
Male ratio =  Average number of male-children per family/Average number of children per family
              =  [2(1/4 +1/8 + 1/8 + 1/16 + 1/16 + 1/16) + 1(1/16 + 1/16 + 1/16 + 1/16) ] / [2(1/4) + 3(1/8 + 1/8)  + 4(1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16)]
              =  [2(11/16) + 1/4] / [2(1/4) + 3(1/4) + 4(1/2)] = [26/16] / [13/4] = 1/2
This implies proportion of females is also 1/2.
Thus, the sex ratio is 1.

[maahi]

can u do this one ??


Let X and Y be two independent random variable with X having a normal
distribution with mean  and variance 1 and Y being the standard normal
distribution.
(a) Find the density of Z = min{X; Y }
(b) For each t belongs  R calculate P(max(X; Y ) - min(X; Y ) > t):

[Sinistral]

oh.. u deleted the first question. I had already solved it. :)
anyways I think u  have already figured out that u need to do convolution.

for the first part of 2nd ques:
P(Z<z)=1-P(Z>z)
P(Z>z)=P(X>z,Y>z)
         =P(X>z)P(Y>z)
         now it becomes a question of calculus. solve it using properties of Gaussian integral.

2nd part:
(I am not sure):
make joint pdf of X & Y
f(x,y)=f(x)f(y)

then assuming X and Y are not equal,
P(max(X; Y ) - min(X; Y ) > t) = P(X-Y>t) + P(X-Y < t)

maahi, are u opting for Option B?

[Sinistral]

Thank you so much.
You make every question look so obvious.

[maahi]

thanks

no

what is d intuition behind this ??  P(X-Y>t) + P(X-Y < t)

[anon_econ]

For part b you need to find the probability that |x-y|>t. Open the modulus sign and try to figure out which distribution x-y follows.