- #1
Maxwell
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(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.
......( 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.