Probability People Come Hither!

[Stokes]

I have this problem which I am having trouble figuring out, can anyone walk me through it?

A survey of families with two children is conducted. The probability of both children being girls is 25%.

(1) Find the probability that at least two out of three families will have two girls. I found the answer to be P(GG) = 1/4 / P(GB,BG,GG) = 3/4 or 1/3 chance of having two girls given the first is a girl. Then I took 1/3 * 1/3 * 2/3 = .074 + 1/3 * 1/3 * 1/3 = .037 for a .11 chance of two out of three families having 2 girls.

I'm not sure if the above is correct.

(2) In a survey of 100 families, find the expected value and standard deviation of the number of families with two girls.

I'm a loss at this one. I know Expected Value = The Sum of (Xi * P(Xi)) and after that I know how to find the standard deviation I just don't know how to figure this one out given the 100 families.

(3) Using the normal approximation to binomial distrubution, find the probability that less than 20 out of 100 families have two girls.

Once again I know binomial mean and Std. Dev is Ux = np and std dev of x = sq rt of (np(1-p)

anyone help me please?

 

[TuxDave]

The probability of both children being girls is 25%.

This is given! Why are you calculating your own conditional probability of having two girls given condition X?

 

[Stokes]

I have a about 20 min before I need to get this is any last minute help?

 

[Atomic Playboy]

You can take my ex's route, and just say the probability of everything is 50/50, cause it either will happen or it won't. But she failed statistics...

 

[sao123]

Originally posted by: TuxDave

The probability of both children being girls is 25%.

This is given! Why are you calculating your own conditional probability of having two girls given condition X?

yea... that is completely unnecessary...




what is the probability of 2 out of 3 families having 2 girls?


The combinations are as follows... Each Y has a 25% chance of occuring, each N has a 75% chance of occuring...

Y, Y, N .25 * .25 * .75 = .046875
Y, N, Y .25 * .25 * .75 = .046875
N, Y, Y .25 * .25 * .75 = .046875
Y, Y, Y .25 * .25 * .25 = .015625
Y, N, N .25 * .75 * .75 = .140625
N, Y, N .25 * .75 * .75 = .140625
N, N, Y .25 * .75 * .75 = .140625
N, N, N .75 * .75 * .75 = .421875

Add up the cases you want to include... = .15625 or 15.625%

Just to check we are doing this right... add them all up and you get 1.



I cant help with 2 & 3... I only had probability... not statistics.

 

[QED]

When in doubt, compute!

Your answer to 1 seems a bit off. In fact, I believe that the probablity is not fixed-- it will vary depending on the number of families included in the survey.

For instance, if there is exactly one family in the survey, then we are essentially asking the probability that it will have both girls, which is of course 1/4.

If there are exactly two families in the survey, then we are asking the probability that both families have two girls, which is simply 1/16.

If there are exactly three families in the survey, then we are asking the probability that at least two of the families have two girls. There are 3 ways that two of the families can have two girls and the third family will not-- each with probablity (1/4)(1/4)(3/4). There is also exactly one way that all 3 families will have two girls, with probability (1/4)(1/4)(1/4). Hence, the desired probability is 3(1/4)(1/4)(3/4) + (1/4)(1/4)(1/4), or 5/32.

Let's double the number of families in the survey to 6... you would think the odds would be the same as when there were 3 families, right? No such luck! In fact, the odds should be less, and here's why: Divide the 6 families into two groups of 3 families apiece (call them Group A and Group B). Now in order for at least two-thirds, or 4, of these 6 families to have two girls each, either Group A has exactly two such families and so does Group B, OR all 3 families in Group A have this property and at least 1 family in Group B does as well (or vice-versa). The odds of the first possibility are 3(1/4)(1/4)(3/4) * 3(1/4)(1/4)(3/4), or 81/4096. The odds of the second possibility are 2 * (1/4)(1/4)(1/4) * ( 1 - (3/4)(3/4)(3/4) ), or 74 / 4096. Hence, the total probablity that at least 4 of the 6 families have two girls is (81+74) / 4096, or 155/4096 (roughly 3.78%).

Using similiar arguments, you can see that the odds will rapidly decrease to 0 as the number of families increases-- and this makes sense. Think of it like this: it's relatively easy for a career .250 batter to hit .666 for a single game, but it's even harder for the same hitter to average .666 for a week, and pretty much impossible to average .666 for a month or a season.

 

[chuckywang]

1) What is the probability that 2 out of 3 families will have 2 girls.

There are 2^6 = 64 total combinations.

You can have:
GG GG GG
GG GG GB
GG GG BB
GG GG BG
GG GB GG
GG BB GG
GG BG GG
BB GG GG
GB GG GG
BG GG GG

That's 10 total ways to have 2 out of 3 familes with 2 girls.

Therefore the probability is 10/64 = 15.625%.

 

[sao123]

and QED and CW both concur with my first answer... so theres 3 diff ways to do the same problem.

 

[sactoking]

Originally posted by: Stokes
I have this problem which I am having trouble figuring out, can anyone walk me through it?

A survey of families with two children is conducted. The probability of both children being girls is 25%.

Hopefully you did not use P(G) = 0.25! Instead it was P(GG) = 0.25. You should have used P(G) = 0.5. If you did, you would have P(GG) =0.25, P(GB) = 0.25, P(BG) = 0.25 and P(BB) = 0.25. In other words, 50% boy, 50% girl, 25% of 2 boys, 25% of two girls, 50% 1 boy and 1 girl.

 

[chuckywang]

2) P(a family will have 2 girls) = 1/4 and P(a family will not have 2 girls) = 3/4


P(x out of 100 families will have 2 girls) = (1/4)^x*(3/4)^(100-x)*C(x,100)
where C(b, a) is a choose b (a combination).

So the expected number of families withi 2 girls is:

Summation of i*(1/4)^i*(3/4)^(100-i)*C(i,100) where i goes from 0 to 100.

There's no easy way to simplify that (I think), so use a calculator.

 

[QED]

Originally posted by: chuckywang
2) P(a family will have 2 girls) = 1/4 and P(a family will not have 2 girls) = 3/4


P(x out of 100 families will have 2 girls) = (1/4)^x*(3/4)^(100-x)*C(x,100)
where C(b, a) is a choose b (a combination).

So the expected number of families withi 2 girls is:

Summation of i*(1/4)^i*(3/4)^(100-i)*C(i,100) where i goes from 0 to 100.

There's no easy way to simplify that (I think), so use a calculator.

Or you can calculate the expected value like this:

For each family there are 3 possibilities: it can have all girls, all boys, or one boy and one girl.

Given 100 families, the sum of the expected values for each of these 3 possibilities has to sum to 100.

So what is the expected value for the number of families who will have one boy and one girl? The odds that a single family will fall into this category is 1/2, and hence the odds that a single family will NOT fall into this category is also 1/2. By symetry then, the expected values for these two possibilities must be equal. They also must sum to 100, so they both are 50.

If a family does not have one boy and one girl, then it must either have two boys, or two girls. By symetry, either of these possibilities is as likely as the other, so their expected values must be equal. They must also sum to 50 (the expected value of the number of families who don't have one boy and one girl). Hence, the expected value, given 100 families, that have two girls is 25.
Among those