(Note -- I use ( x ) to mean binomial coefficients. I'm not sure how to(adsbygoogle = window.adsbygoogle || []).push({});

....................( 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.

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# Homework Help: HW Question (Probability)

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