Don't know if this is "efficient" but I think it works...
yn <- c("Y", "N")
X <- expand.grid(x1=yn, x2=yn, x3=yn, x4=yn)
Yp <- c(0.6, 0.5, 0.8, 0.9)
X$prob <- apply(X, 1, function(x) cumprod(ifelse(x == "Y", Yp,
1-Yp))[length(x)])
Michael
On 9 November 2010 17:05, Kate Hsu wrote:
> Dear r user
Dear r users,
I have 4 variables x1,x2,x3,x4 and each one has two levels, for example Y
and N.
For x1: prob(Y)=0.6, prob(N)=0.4;
For x2: prob(Y)=0.5, prob(N)=0.5;
For x3: prob(Y)=0.8, prob(N)=0.2;
For x4: prob(Y)=0.9, prob(N)=0.1;
Therefore, the sample space for (x1, x2, x3, x4)={, YYYN, Y
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