Probability Calculations for Disease and Card Decks

[Maxwell]

(Note -- I use ( x ) to mean binomial coefficients. I'm not sure how to
......( y )
get them right in LaTeX and I'd rather spend time working on my problems for now.)

Suppose that the chance of a special disease in human births is 0.002. (1) What is the probability of observing at least 4 (including 4) infants having the disease in 1000 human births?
(2) How many infants are expected to have the disease and what is the variance?

Ok, for (1) I have this:

N=1000, j=4, p = disease = 0.002

So I used the binomial formula:

( N )
( j ) * (p)^j * (1-p)^(N-j)

( 1000 )
( 4 ) * (0.002)^4 * (1-0.002)^(1000-4)

Is this right?

For (2): I have no idea...


For a deck of well shuffled cards,
(1) How many Aces do you expect to appear among the top 6 cards?
(2) What is the chance that the top 6 cards are made of 3 Kings, 2 Queens, and 1 Ace?

For (1): I thought to use the Hypergeometric formula.

N = 52, n = 6, M = # of aces = 4, j = ??

Hypergeo:

( M ) ( N-M )
( j ) ( n-j )
-------------
( N )
( n )

( 4 ) ( 48 )
( j ) ( 6-j )
-------------
( 52 )
( 6 )

For (2): I used the Hypergeometric form again.

N = 52, n = 6,
M1 = # of Kings = 4,
M2 = # of Queens = 4,
M3 = # of Aces = 4,
j1 = 3, j2 = 2, j3 = 1

( M1 )( M2 )( M3 )( N-M1-M2-M3 )
( j1 )( j2 )( j3 )( n-j1-j2-j3 )
--------------------------
( N )
( n )

( 4 )( 4 )( 4 )( 40 )*
( 3 )( 2 )( 1 )( 0 )**
-------------------
( 52 )
( 6 )

* N-M1-M2-M3 = 40
** n - j1 -j2 -j3 = 0


Thanks for the help and I hope this post is understood.

 

[0rthodontist]

In the first one, you've given the probability that _exactly_ four babies have the disease. The probability that _at least_ four babies have the disease is 1 - (the probability that 0, 1, 2, or 3 babies have the disease). For the second part, how is the disease incidence distributed? You're just looking for the mean and variance of that distribution.

In the second one, part 1, you are on the right track with the hypergeometric distribution and you've set it up basically right but what are they asking? They are not asking the probability that you will draw any particular number of aces, they are asking the mean number of aces. The second part is correct.

[tex]a\choose b[/tex]

 

[Maxwell]

Thanks, 0rthodontist, this helps a lot.

For the first problem, second part, we are looking for the mean and variance. I'm not really sure how to set up the equations using the given variables.

mean = E{x} = mu1{x} = (0.002)*SUM(i=0 to 1000) of ??

What exactly would I be summing?

If I find that out, I can find variance by V{x} = mu2{x} - mu1{x}^2 , where mu2{x} = E{x^2}

The second problem I understand now.

Thanks again.

 

[0rthodontist]

I don't think you're really expected to find the mean and variance on your own--you can just look it up. The disease is distributed according to the same distribution you used to find the probability of any particular number of babies having the disease.

 

[Maxwell]

I see. Thanks a bunch for your help, Orthodontist.