I'm having trouble with AT LEAST in combinatorics. we have 4

Assuming no restrictions (i.e. can repeat the digits).

P(At least 1 Prime) : 1 - P(No Prime)

p(no prime) : 2^-4 = 1/16 ---> selecting 4,6,8,or 9 four times

so 1- (1/16) = 15/16

P(At least 2 Prime) : 1 - [P(No Prime) + P(1 prime)]

p(1 prime) = C(4,1) * (1/2)*(1/2)*(1/2)*(1/2) = 4* 2^-4 = 1/4
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8

so 1 - [(1/16) + (1/4)] = 11/16

P(At least 3 Prime) : 1 - [P(No Prime) + P(1 prime)+ P(2 prime)]
p(2 prime) = C(4,2) * (1/2)*(1/2)*(1/2)*(1/2) = 6* 2^-4 = 3/8
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8

so 1 - [(1/16) + (1/4) + (3/8)] = 5/16

If there's a shorter way, i'd like to know. I think it gets much more confusing when there are restrictions such as "exactly 2 primes together".

For instance, what is the probability of getting 3 primes, two of which are back to back - e.g. 2385