Originally Posted by

**matheagle** I would treat this as a multinomial with 3 groups.

I would add the probability of exactly 1, exactly 2, exactly 3 or exactly 4 times we select a 6 as the highest number.

$\displaystyle {4 \choose 3,1,0}(.6)^3(.1)^1(.3)^0+{4 \choose 2,2,0}(.6)^2(.1)^2(.3)^0$

$\displaystyle +{4 \choose 1,3,0}(.6)^1(.1)^3(.3)^0+{4\choose 0,4,0}(.6)^0(.1)^4(.3)^0$

$\displaystyle =.0864+.0216+.0024+.0001=.1105$

The $\displaystyle [(7^4)-(6^4)]/10^4$ in the different is the two sets.

If you list all possible sets of four numbers and you first select any number from 0 to 6, but then you remove

all of those that only consist of 0 through 5 you end up with at least one 6 somewhere.